iit – jee: 2017 assignment topic: rotational dynamics 1. · iit – jee: 2017 assignment topic:...

CENTERS: MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 1

ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI IIT – JEE: 2017 ASSIGNMENT

TOPIC: ROTATIONAL DYNAMICS

SOLUTION

1. (a) since the external forces acting on the system are all vertical, the centre of mass of the system

will not move horizontally. When the ball has left the hemisphere, it must move vertically upwards: since the ball slides all the way along the hemisphere, it cannot have a horizontal velocity component at separation. Otherwise, its horizontal velocity would be equal to that of the cart, which would mean a displacement of the centre of mass of the system to the left or to the right. After separation, the motion of the ball is vertical projection, and the cart is brought to rest again. Thus, all the energy is concentrated on the ball again, and the ball will rise back to its initial height.

Let X denote the displacement of the cart and let x denote the horizontal component of the displacement of the ball. As shown in the figure, the distance of the centre of mass of the cart from the common centre of mass is 2X , and the distance of the ball from there is 2R X metres. By definition of the centre of mass,

2 2X XM m R

,

and hence the displacement of the cart is

2mX RM m

To the right, while horizontal displacement of the ball is

2Mx RM m

Toe the left. With numerical data:

2.2kg .0.5m 0.8m2kg 0.5kg

x

.

(b) The normal force acting at the lowermost point of the path of the ball is obtained by applying Newton’s second law:

2relN mg mR

, (1)

Where N is the normal force and rel is the speed of the ball relative to the cart. At that time instant the cart represents an inertial reference frame, therefore Newton’s law is valid in it as well.

From the work-energy theorem,

2 21 12 2

gm h R m MV , (2)

Where and V are the speeds of the ball and the cart, respectively, relative to the ground. Since horizontal momentum is conserved, m MV (3) Finally, the relationship of the speeds measured in the two reference frames is rel V The solution of the simultaneous equations for the magnitudes of the quantities involved is

2

2

2m g h RV

M mM

and 2Mg h RM Vm M m

,


2 2

max 2

2kg 0.5kg 0.25kg 0.5m 0.5m m2 2 0.5m 10 30 N2kg 0.5m s

Mm m h RN m gM R

.

2. (a) When point A hits the ground, the rods form a straight

line. As rods have constant length, the two ends of the system must have zero velocity at that moment, so the block of mass

2m stops. Since only conservative forces act in this situation, the initial gravitational potential energy of the rods will be transformed into the rotational kinetic energy of the rods, which is independent of the mass of the block.

At the moment when point A hits the ground, the rotational kinetic energies of the two rods are equal. Assuming that the rotational inertia of a rod about its end is 2

1 / 3m L , the rotational kinetic energy of one rod is :

2 2 21 1

1 1 12 2 3

E m L .

The law of conservation of energy will take the from of:

2 21 1

1 12 sin 22 2 3Lm g m L ,

Where / 2 sinL is the distance of the centre of mass of each rod from the ground in their initial position. As o60 , we know that . Substituting this into the equation and isolating the angular velocity, we find:

3 32g

L .

We then use this to express the velocity of point A when hitting the ground:

23 3 3 9.81ms 0.5m 3 m3.57

2 2 sAgLL

Note that the result does not depend on the masses of either rod or the block.

(b) The acceleration of mass 2m at the moment when point A hits the ground is made up of two

different components: the first one is the horizontal component of the acceleration of point xA a , the second one is the horizontal component of the acceleration of point B relative to point A.

Let the reference frame be attached to point B, whose velocity becomes zero, when point A hits the ground. In this frame rod AB (and therefore point A) rotates in an accelerated motion around point B. The horizontal component of the acceleration of point A must be the centripetal acceleration, so

2x n Aa a L

If the reference frame is now attached to point A, the horizontal acceleration of point B (and therefore that of mass 2m ) will have the same magnitude as the one calculated above but will be in opposite


directions. As the horizontal acceleration of point A in this situation has magnitude 2x n Aa a L

and points towards point O, therefore the total acceleration of point B and mass 2m at the moment when 0B is :

22 2B xa a L Substituting known values gives:

2

3 3 m2 2 3 3 50.972 sB x

g La a gL

Note that this result is independent of the masses and of the lengths of the rods. (The force acting in

hinge B is 2 2

m2kg 50.97 101.94 NsBF m a .)

3. The figure shows that the vertical plane containing the rod is perpendicular to the axis of the cylinder, so the rod will move in that plane.

The change in gravitational potential energy can be determined by comparing the initial and final states. Since the system is conservative, the opposite of that change equals the total kinetic energy gained.

When point B reaches the cylinder, the rod and the radius drawn to it are perpendicular to each other. As seen in the figure, they enclose a triangle with sides

, 2OB R BA R

2 24 5OA R R R .

The distance of the centre of mass S of the rod from the horizontal plane in the initial and final

positions are

12

2h R and 2

55

h R ,

Since 2 : : 5h R R R Energy is conserved:

2 21 2

1 12 2s smg h h m .


The instantaneous sped s of the centre of mass S is determind by the instantaneous angular

speed and the distance Br CB of instantaneous axs of rotation C of the rod from the point S. Since the velocity of point A is horizontal and that of point B – at that time instant – is tangential to the cylinder, the instantaneous axis of rotation is the intersection of the perpendiculars drawn to these velocities. From geometry, the instantaneous radius Ar CA drawn to point A satisfies

: 5 2 :rlA R R R , thus 2 5Ar R . The instantaneous radius of rotation of point B is obtained from the Pythegorean theorem: 2 2 2 24 20 4 4B Ar r R R R R .

Finally, the instantaneous radius Sr CS at the centre of mass is

2 2 2 216 17Sr CB CS R R R .

Since the speed to be found is B Br , the angular speed remains to be determined. The energy equation in terms of the unknown angular speed and the radius Sr obtained above is

2 2 2 2 2 22 5 1 1 1 2617 42 5 2 2 12 3

mfR m R m R mR

.

The solution of the equation for is

13 2 5 0.546s26 2 5

gR

Thus the speed of the point B of the rod at the time instant in questions is

m4 2.19sB R .

4. Let A be the topmost point of the semi-cylinder. The length of the cart is a minimum if it accelerates

to the minimum possible speed V during its contact with the small object, and this covers the shortest possible distance during the time of the fall.

While the two objects are in contact, the cart keeps on acceleration, with the single exception of the time instant when the small object reaches the height of 2R on the semicircle.

The force exerted by the small object on the cart is the smallest if it slides onto the cart at the minimum possible speed. Since the condition for reaching the top is that the normal force N exerted by the track should remains 0N all the way, the cart will gain the smallest possible final speed if the normal force N decreases to 0 exactly at the topmost point. Let us go through the conditions required.

Condition 1: 0AN (1) When the object has reached point A, the cart moves on with uniform motion in a straight line, thus it

becomes an inertial reference frame. In the reference frame of the cart, the small object moves along a circular path of radius R until it reaches A-ig, and then continues on a parabolic path as a projectile with a horizontal initial velocity. (In the reference frame attached to the ground, it reaches the cart with free fall along a vertical line, with zero initial velocity.) The next condition is provided by the circular motion in the reference frame of the cart:

Condition 2:

2rel

Amg N mR

,

and thus with (1),


2rel .gR

(2)

Since there is no friction, all forces are conservative, the total mechanical energy of the system is conserved. Let 0 denote the speed of the small object when it hits the cart (same as its initial speed of sliding onto the cart).

Condition 3:

2 20 1

1 1 1 22 2 2

m m MVh mgh ,

Where v is the speed of the cart and 1 is the speed of the small object acquired by the time the small object rises to a height h above the surface of the cart. The small object needs to reach the point A at a height of 2h R , and it needs to lose its speed there ( 1 0 is needed). Multiplied by two, the equation takes the form

2 20 4m MV mgR (3)

Since the small object is to stop at the topmost point of its path, its velocity relative to the cart at that point is the opposite of the velocity of the cart. In absolute value,

rel V (4) Since the external forces are all vertical, the sum of horizontal momenta is constant, that is, Condition 4: 0mv MV . (5)

From (2) and (4): V gR . (6) (6) is needed for answering question a) since the distance covered by the cart can be determined

with (6) from the time of the fall from a height of 2R. The distance covered in free fall is

212

2R gt ,

And hence the time of fall is

4Rtg

.

In that time interval, the cart covers a distance of

24. 4 2Rs Vt gR R Rg

.

Therefore the minimum cart length required is min 2L s , that is, min 4 4.0.36 1.44L R m m . b) The equations for momentum (5) and energy (3) both have to be satisfied. The mass ratio of the

small mass to the cart is not arbitrary. From (5) and (6),

0 .M MV gRm m

(7)

Substituted into (3):

2

2 4 .Mm gR M gR mgRm

With both sides divided by m:

2

2 4 ,M MgR gR gRm m

and hence, the quadratic equation


2

2 4 0M Mm m

is obtained for the mass ratio. With the notation 2/ : 4 0M m k k k . Hence

1 1 16 2.56 (15),2

k

that is, 2.56M m , and since 1m kg , the mass of the cart is 2.56M kg , Independently of R. c) Since there is no friction, the small object was released at the height of

20 ,

2h

g

Which can be calculated with the use of (7):

2

21 1 2.56 0.36m 1.18(1)m2 2

Mh Rm

.

The small object slides onto the cart with a speed of

0 22.56 9.8 0.36 4.85 7 ,M m mgRm s s

And the final speed of the cart is

29.8 0.36 1.89 6m mV gR ms s

.

5. Although the setting of the problem doesn’t state it exactly, let us assume that the disk is

homogenous and rotates around a vertical axis that goes through its centre. It is also important to state that the disk is much greater than the man, therefore the latter can be handled as a pointmass, which means that the man’s rotational inertia when standing in the centre will be taken as zero.

Since there is no external torque acting on the system (friction and air resistance are neglected), the total angular momentum of the system remains unchanged. The total angular momentum of the system is the sum of the angular momentum of the disk disk and the moment of momentum of

the man manm R . Applying the conservation of angular momentum:

2disk 1 man 1 disk 2 0m R ,

Since (as it was stated above) the man’s rotational inertia is zero at the centre. Isolating the final angular velocity, we find:

2

disk man2 1

disk

m R

.

The change in the total energy of the system is:

2 2 2 22 disk 2 disk 1 man 1

1 1 112 2 2

E E m R

.

After substituting the expression for 2 and rearranging the equation, we have:

2 2

2 2 2 2 2disk man disk manman 1 man

disk disk

42 2

m R m RE m R m R n

.

Using that the rotational inertia of a homogeneous disk about its axis of rotational symmetry is 21

2disk diskm R , the change in the energy can be written as:

2 2 22 42

disk manman

disk

m mE m R nm

,


Substituting values gives: 2 2 200 160 80 25 4 0.01 605.3J.600

E kg m s

The change in the kinetic energy of the system is 605.3 J, which means that the total energy of the system does not remain constant. This of course doesn’t mean that the law of conservation of energy is violated, because the increase in the energy of the system is caused by the work done by the man’s muscled, so biological (chemical) energy is transferred to kinetic energy in this case.

6. Let h and , respectively, denote the magnitudes of the horizontal and vertical velocity

components of the sliding object. Let m stand for the mass of the object, M for the mass of the cylinder, and let and denote the angular velocity and moment of inertia of cylinder about its axis.

a) Mechanical energy is conserved:

2 2 21 12 2hmgh m . (1)

Angular momentum is conserved. With respect to the axis of the cylinder, the sum of the angular momenta of the moving object and the cylinder is zero:

0hm R (2) In order to determine the three unknowns , ,h , a third equation is needed. It will be provided

by the condition that the object remains on the track. Note that while the object descends through a height of h, the endpoint of the horizontal radius drawn to its position turns through a distance of

1R in a horizontal direction, while the circumference of the cylinder rotates through 2R in the opposite direction. The slope of the track tan / 2h R can be expressed in terms of these quantities:

1 2

1

12tan 1 1 2

2 2h

th hR R RR t R t

, (3)

Since the initial speed was zero and acceleration is constant. The system of equations (1)-(2)-(3) becomes simpler with the use of the relationship 5M m of the

masses and the formula 212

M R for the rotational inertia of the cylinder:

2 2 2 22 2.5hmgh m mR 1'

22.5hm R mR 2'

2h

hRR

. 3'

With 1' and 2' divided through by m, and 2' also divided by R,

2 2 2 22 2.5hgh R 1"

2.5h R 2"

.2h

hRR

3'

Rcan be expressed from 2" and substituted into 1" and 3' :

2 2 2

2 22

3.5 2.52 2.52.5 2.5

h hhgh

,

That is, 2 25 3.5 2.5hgh , (4) And


3.52.5 2 5

hh h

h hR R

(5)


7.


8. Let a be the acceleration of mass m. According to Newton’s second law, the tension in the string is; K m g a .


9.

10.


11. First solution


Second solution


12. Let iF be the friction force between the stick and the ith disk, with I = 1 corresponding to the leftmost disk. Let a be the acceleration of the stick. Then the non-slipping condition says that the angular acceleration of the ith disk is / i ia r . So 1 applied to the ith disk gives

2

1

1 1 .2 2

i i i i i i i iaI F r m r F m ar

Since the mass of each disk is proportional to its area, which in turn is proportional to its radius squared, the im are given by 1 2, / 9 m m m m , 3 / 81m m ,etc. The F ma equation for the stick, along its direction of motion, is (using / 2i iF m a in the second line below) 1 2sin30 ... omg F F ma

1 1 1 11 1 ...2 2 9 81

mg ma

1 112 2 1 1/ 9

g a

25 82 16 25

g ga a .

Remark: This acceleration is independent of R (and also m); this follows from dimensional analysis, as you can verify, because the given parameters are g, R, and m. So as long as the mass of the stick equals the mass of the leftmost disk, and as long as the radii of the disks decrease by successive factors of 1/3, the system can be built to any size and the acceleration of the stick will always be 8g/ 25. 13. In fig. , there are five unknowns: f , , , , c sF T and .ca So we need five equations: I for the cylinder, relative to the center:

2

f 2

cmRF R T R . …(1)

I for the stick, relative to the pivot:

22sin 2

3

s

m Rmg R T R . …(2)

F ma for the cylinder, along the plane: fsin . cmg T F ma …(3) Non-slipping condition for the cylinder: c ca R . …(4) Conservation-of-string equation: c sa …(5) This can be seen from the following reasoning. Because of the string, the linear accelerations of the top of the stick and the “top” of the cylinder must be equal. The first of these is 2 sR . The second is c ca R , because this is the acceleration of the center of the cylinder plus the acceleration of the rim relative to the center. But c ca R Eq.(4) , so we end up with 2 cR for the acceleration of the


“top” of the cylinder. Hence 2 2 s c s cR R . The intuitive reason for this is that the cylinder rotates around its bottom point (which is instantaneously at rest), just as the stick does. From Eqs. (4) and (5), both ’s are equal to /ca R . So we are now down to three unknowns f , , cF T a and three equations (the torque and force equations). These equations become, respectively,

f1 ,2

cF T ma

4sin 2 ,3

cmg T ma

fsin . cmg T F ma …(6) Adding these three equations conveniently eliminates both T the fF , and we obtain

17 12 sin2 sin6 17

c c

gmg ma a …(7)

Note that we never needed to use the F ma equation for the stick. That would serve only to yield the force at the pivot. 14. (a) There are no external forces, so p and L are conserved. E might or might not be; we’ll find out eventually. Conservation of p quickly gives the speed of the dumbbell’s CM as v/2, although we won’t need this to find . The CM is halfway between the masses, which means / 2 from each. The moment of inertia of the resulting dumbbell is therefore 2 22 / 2 / 2 I m m . So conservation of L around the point on the table that coincides with the CM at the moment of the collision gives (using the fact that the moving mass has an “impact parameter” of / 2 2

2

0 ,22 2 2

m vmv …(1)

Where the zero here comes from the fact that the CM is moving directly away from the origin, so the first term in Eq. is zero. (b) The energy of the dumbbell is partly translational and partly rotational. So the energy lost to heat (if any) is

2 2 2i f

1 1 122 2 2

CME E mv m v I

22 2

21 1 122 2 2 2 2 2

v m vmv m

2 21 1 1 12 4 8 8

mv mv …(2)

This is nonzero, so some energy is indeed lost to heat. (c) The velocity of the lower-right mass is the sum of the CM velocity plus the rotational velocity relative to the CM. The former is / 2v to the right. The latter is / 2 / 2 / 2 2 r v v down

and to the left at a 45o angle (since the rotation is clockwise). The sum of these two vectors is shown in fig. The desired velocity is therefore , / 4, / 4 x yv v v v . Equivalently, it points down to the

right at 45o , with magnitude / 2 2v .


15. Energy is conserved during the process, because we are told that all interactions (balls with stick, and stick with table) are elastic. Additionally, the angular momentum of the system relative to the corner of the table is conserved, because the external force from the corner provides no torque around it. (During the collision, the table applies a force to the stick only at the corner.) Let 1v and 2v be the velocities of the two balls right after the collision, with the positive directions chosen as shown in fig. If 1v comes out to be negative, that means the left ball moves upward right after the collision. With counterclockwise taken to be positive, conservation of angular momentum relative to the corner of the table gives

0 1 2 1 0 2/ mv a mv a mv b v v b a v . …(1) And conservation of energy gives

2 2 2 2 2 20 1 2 0 1 2

1 1 12 2 2

mv mv mv v v v . …(2)

Plugging the 1v from eq. (1) into Eq. (2) gives

22 2 2 2 20 0 2 2 0 2 2/ 0 2 / / 1 v v b a v v v v b a v b a

2 0 2 2

2

abv v

a b. …(3)

Either of Eq. (1) or Eq. (2) then gives

2 2

1 0 2 2

a bv va b

. …(4)

Limits: If b0 (more precisely, if b << a) then 1 0v v and 2 0v . So the left ball keeps heading downward at speed 0v , and the right ball doesn’t move. This makes sense intuitively. If a0 (more precisely, if a << b) then 1 0 v v and 2 0v . So the left ball bounces upward (due to the minus sign in 1v ) with the same speed 0v that it originally had, and the right ball doesn’t move. This also makes sense intuitively; the left ball is essentially just bouncing off the table. If a b then 1 0v and 2 0v v . So the left ball ends up at rest (instantaneously), and the right ball moves upward with speed 0v . It is believable, although not obvious, that a = b leads to these velocities. However, by a continuity argument, it is obvious that there must be some relation between a and b that leads to the final speed of the left ball being zero, because 1v is positive (downward) if b << a , and it is negative (upward) if a << b , so it must be zero for some intermediate value. By conservation of energy, 1v can’t be any larger than 0v , of course. So the a = b case yields the maximum possible speed of the right mass. (You can also deduce this form Eq. (3). If the masses of the balls aren’t equal, then the maximum possible speed of the right ball can quickly be found by using conservation of energy and letting the final speed of the left ball be zero; the right ball’s final kinetic energy equals the left ball’s initial kinetic energy. This method of finding the maximum speed of the right ball is much cleaner than finding the right ball’s final speed and then taking the derivative to maximize it.


17.


18.


Taking the derivative of this, we find that the force from the wall is maximum when

1 19cos 26.7 .6

o

iit – jee: 2017 assignment topic: rotational dynamics 1. · iit – jee: 2017 assignment topic:...

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