PHY-2048C – Spring 2021 SI with Camilo

Exam 4 Practice Test

Disclaimer: This practice test does not necessarily cover all the material that is going to appear on the test, the questions are not the same as the actual test, and should not be used as the sole study guide for the test.

1. Starting from rest, a 15-cm-diameter compact disk takes 4.0 s to reach an angular velocity of 3000 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia 3x10-5kgm2.

a. How much torque is applied to the disk?

b. How many revolutions does it make before reaching full speed?

ωf=3000 rpm=3000 rev /m

2 πrad1rev

∗1m

60 s=314.16 rad / s

R = 0.075

α=ωf

t=314.16 rad / s

4 s=78.54 rad / s2

∑ τ=αI=(78.54 )∗3 x10−5=0.00236Nm

θ=ω0 t+12α∗t2

θ=12(78.54)∗42=628.319 rad

628.319 rad=628.319 rad 1 rev2 πrad

=100 rev

2. Two masses are attached to a massless string that goes around a massive pulley as shown in the picture. Find an equation that gives the value of the acceleration of the system (a) in terms of m1, m2, M, and g.

3. Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods.

a. Find the rotational inertia of this system about an axis that coincides with one side.

I a=m L2+mL2=2mL2

b. Find the rotational inertia of this system about an axis that bisects two opposite sides.

I b=4m(L/2)2=mL2

4. A typical small rescue helicopter has four blades: Each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. 

(a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm.

∑ I=4∗( 13mblade(r blade)

2)=4∗( 13∗50∗(4)2)=1066.667 kg (m)2

ω=

300 revmin

∗1min

60 sec∗2 πrad

1 rev=10πrad / s

KEr=12I ω2=0.5∗1066.667∗(10 π2 )=526378.9 J

5. A 25-kg child stands at a distance r=1.0m from the axis of a rotating merry-go-round. The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.(treat the child as a point mass)

I total=I child+ Imerry=mchild rchild2+ 12mmerry rmerry

2

I total=25∗12+0.5∗500∗22=1025 kgm2

6. Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.

7. Find the expression for the moment of inertia of a thin rod with length L if the axis of rotation is on the middle of the rod.

8. A 2.2 kg block rests on a 30 ∘ slope and is attached by a string of negligible mass to a solid drum of mass 0.5 kg and radius 6.3 cm , as shown in the figure (Figure 1) . When released, the block accelerates down the slope at 1.8 m/(s^2) .What is the coefficient of friction between block and slope?

m=2.2 kgθ=30M=0.5kgr=0.063ma=1.8m /s 2

9. A rod of length L = 1.0 m and mass M = 2 kg is hinged on one side so that it can rotate freely(imagine like a door but it is a rod). A bullet is then fired and hits the rod right at the edge. The bullet has mass m=5g and a speed of 200 m/s. What is the angular velocity of the rod with respect to the hinge just after the bullets embeds itself in the door?

m = mass bullet M = mass of rod

L0=LF

L=Iω

L0=Lbullet 0+Ldoor 0

Ldoor0=0

L0=I bullet❑ωbullet❑=m R2ω=mR(Rω)=mRv

Rω=v

LF=Lbullet∧door f=I ω=(mR2+ 1

3M R2)ωf

mRv=(mR2+ 13M R2)ωf

mRv

(m R2+ 13M R2)

=ωf=(0.005)∗200

(0.005+ 132)

=1.5rad /s

10. A flywheel rotates without friction at an angular velocity ω_0=600rev/min on a frictionless, vertical shaft of negligible rotational inertia. A second flywheel, which is at rest and has a moment of inertia three times that of the rotating flywheel, is dropped onto it. Because friction exists between the surfaces, the flywheels very quickly reach the same rotational velocity, after which they spin together. (a) Use the law of conservation of angular momentum to determine the angular velocity ω of the combination.

11. You are headwaiter at a new restaurant, and your boss asks you to hang a sign for her. You’re to hang the sign, whose mass is 55 kg, in the configuration shown in the figure. A uniform horizontal rod of mass 7.3 kg and length 2.3 m holds the sign. At one end the rod is attached to the wall by a pivot; at the other end it’s supported by a cable that can withstand a maximum tension of 760 N.You’re to determine the minimum height hmin above the pivot for anchoring the cable to the wall.

∑ τ=0=τ T−τg=T∗L∗sin (θ)−0.5∗L∗Fg

T∗L∗sin(θ)=0.5∗L∗F g=0.5∗L∗mt∗g

θ=arcsin (0.5∗mt∗g

T)=26.09°

tan(θ)= h2.3

h=2.3∗tan(θ)=1.1m

12. A uniform ladder is L=5.0m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in the figure. The inclination angle between the ladder and the rough floor is β=53°. Find the reaction force from the wall on the ladder and the coefficient of static friction μs at the interface of the ladder with the floor that prevents the ladder from slipping.

Fnety=N floor−W=0

N floor=W=400 N

Fnetx=F f−Nwall=0

Nwall=F f=μsN floor

I decide to let the axis of rotation to be on the lower end of the ladder to make the torques of the forces from floor equal 0

τ net=τweight−τnormal force of thewall=400(2.5)sin(37)−N wall(5)sin(53)=0

Nwall=150.7N

Nwall=μsN floor

μs=Nwall

N floor=0.377

13. Determine the semi-major axis of the orbit of Halley’s comet, given that it arrives at perihelion every 75.3 years. If the perihelion is 0.586 AU, what is the aphelion?

14. Suppose a small planet is discovered that is 14 times as far from the sun as the Earth's distance is from the sun (1.5 x 1011 m). Use Kepler's law of harmonies to predict the orbital period of such a planet. (Assuming circular orbits)

Using

(T E)2

(RE)3=

(T P)2

(RP)3

(T P)2=

(T E)2∗(RP)

3

(RE)3 =(T E)

2(RP

RE)3

=(1 year )2(14)3=2744 yr 2

T P=52.383 yr