Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

PHYSICS XXXXXXXXXXPRACTICE EXAM #3 1. A solid cylinder is pivoted about a frictionless axle as shown. A rope wrapped around the outer radius of 2 m exerts a downward force of 3 N. A rope wrapped...

1 answer below »
PHYSICS XXXXXXXXXXPRACTICE EXAM #3
1. A solid cylinder is pivoted about a frictionless axle as shown.
A rope wrapped around the outer radius of 2 m exerts a downward
force of 3 N. A rope wrapped around the inner radius of 0.7 m
exerts a force of 8 N to the right. The moment of inertia of
the cylinder is 8 kg m . Find the angular acceleration. 2
2. If the cylinder in problem 1 is initially at rest, how long will
it take for the cylinder to turn one revolution?
3. A uniform rod of mass M and length L is free to rotate about a pivot
at the left end. It is released from rest in the horizontal
position (q = 90 ). What is the torque on the rod when it makes an o
angle q with the vertical?
4. In problem 3, what is the downward linear acceleration of the right end of
the rod when it is first released (at q = 90 )?
o
5. Find the kinetic energy of a solid sphere of mass 0.5 kg and radius 10 cm
that rolls without slipping on level ground at 12 m/s.
6. If the sphere in problem 5 rolls up a hill, how far above the ground will the
sphere climb before it rolls back down?
7. A solid disk of radius 5 m and mass 8 kg rotates clockwise at
1.5 rad/s. Above this disk is a hoop of radius 2.5 m and mass
8 kg, rotating counterclockwise at 3 rad/s. The hoop drops down
onto the disk, and friction causes them to rotate together.
Find the final angular velocity.
8. In problem 7, find the kinetic energy lost in this collision.
9. A man tries to raise a 75 kg flagpole that is attached to the
ground by a frictionless pivot. The pole is 6 m long. The
man pulls on a rope attached to the top of the pole with a
force of 255 N. Find the net torque on the flagpole.
10. Find the angular acceleration of the flagpole in problem 9.
11. The man in problem 9 changes his force so the pole is at rest in static
equilibrium. Find the tension in the rope, and the horizontal and vertical
forces at the flagpole’s pivot.
12. A 1.5 kg frog sits at rest on top of a solid disk. The
disk's mass is 4 kg, its radius is 1.25 m, and it rotates
on a frictionless axle. The frog jumps from the disk at
3.7 m/s at an angle of 50 . Find the angular momentum of o
the frog about the axle as it leaves the disk.
13. How much time will it take the disk in problem 12 to rotate one complete
revolution after the frog jumps?
14. A sailor on a lake uses a rope to lower an iron ball of radius 30 cm to a
depth of 80 m below the surface of the water. What is the pressure (in atm)
at that depth? The density of water is 1000 kg/m .
3
15. Find the tension in the rope in the previous problem. The density of iron
is 7.86 x 10 kg/m .
3 3
16. A hollow steel ball has a radius of 1.5 m and a mass of 15 kg. Inside the
ball is a vacuum. The ball is anchored to the ground by a cord. Find the
tension in the cord. The density of air is 1.29 kg/m . 3
17. Water flows through a horizontal pipe. At position 1 (the wide end of the
pipe) the pressure is 3 x 10 Pa and the water speed is 12 m/s. At
5
position 2 (the narrow end), the pressure is 1.3 x 105 Pa. What is the
water speed at position 2?
18. The diameter of the pipe in problem 14 is 65 cm at position 1. What is the
diameter of the pipe at position 2?
19. The water from the pipe in problem 19 flows into a tank of volume 180 m . If
3
the tank is initially empty, how long will it take to fill the tank?
20. A satellite has a mass of 100 kg and is at an altitude of 2 x 10 m above the
6
ground. What is the potential energy of the satellite-Earth system?
21. In problem 20, what is the force of gravity on the satellite?
22. A white dwarf is a compact star. Its mass is equal to that of the Sun, but
its radius is that of Earth. Find the acceleration due to gravity at a white
dwarf's surface.
23. An apple is dropped from a height of 12.8 x 10 m above the surface of the
6
white dwarf described in problem 22. With what speed does the apple strike
the surface of the white dwarf?
24. What is the minimum energy needed to send a 3000 kg spacecraft from Earth to
an infinitely distant point in space?
25. A neutron star is very compact. If the escape velocity of a neutron star
with the same mass as the Sun is 1.5 x 10 m/s (half the speed of light), what
8
is the radius of the neutron star?
26. When a mass is attached to the end of a vertical spring, the spring is
stretched down 3 cm. If the mass is pulled down a bit farther and then
released, what is the period of oscillation of the mass on the spring?
27. In problem 26, if the mass moves through its equilibrium position at 50 cm/s,
what is the amplitude of the oscillation?
28. On Mars, a pendulum of length 2.36 m swings back and forth once in 5 seconds.
What is the acceleration due to gravity on Mars?
29. A Christmas tree ball hangs from a hook in the ceiling. If the radius of the
ball is 8 cm, find the period of the ball's oscillation as it swings on the
hook.
Answers: XXXXXXXXXXrad/s XXXXXXXXXXsec (3) -(Lmg sin q)/ XXXXXXXXXXm/s
2 2
XXXXXXXXXXJ XXXXXXXXXXm XXXXXXXXXXJ XXXXXXXXXXN m XXXXXXXXXXrad/s ,
2
h v counterclockwise (11) T = 220 N, F = 206 N, F = 810 N XXXXXXXXXXkg m /s,
2
clockwise XXXXXXXXXXs XXXXXXXXXXatm XXXXXXXXXXN XXXXXXXXXXN XXXXXXXXXXm/s
XXXXXXXXXXcm XXXXXXXXXXs XXXXXXXXXXx 10 J XXXXXXXXXXN XXXXXXXXXXx 10 m/s 9 6 2
XXXXXXXXXXx 10 m/s XXXXXXXXXXx 10 J XXXXXXXXXXkm XXXXXXXXXXs XXXXXXXXXXcm 6 11
XXXXXXXXXXm/s XXXXXXXXXXs
2
005_5k-fhiu0ojx.pdf 005_5k-swzja14k.pdf
Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
119 Votes
Microsoft Word - Solution.docx
1. Given that moment of inertia of the cylinder is 8kgm2

In this case we have two torques acting , the 1Ï„ , will try to rotate the cylinder in clockwise direction about
the axle and 2Ï„ will try to rotate the cylinder in anticlockwise direction about the axle.

∴ the net torque 1 2( ) −τ = τ τ
= f1r1 – f2r2 ( ∵ torque = force X ⊥ distance )
= 3x2 -8x0.7
= 6 – 5.6
= 0.4 Nm
this net torque will produce an angular acceleration given by
Iτ = ω

Where I = moment of inertia of cylinder
α = angular acceleration

∴ α =
I
Ï„
=
0.4
8
= 0.05 radsec
-2
2. The angular displacement is given by
2
o o
1
t t
2
θ = θ + ω + α
Here o o0and t 0θ = ω = , because it starts from rest (as given )
∴ 2
1
t
2
θ = α
We have to find the time for one revolution
ie 2θ = π
212 t
2
4
t
∴ π = α
Ï€
=
α

after, substituting the numerical values in the above equation , we get
t = 15.85 sec
t = 15.9 sec
3
The Centre of gravity of a rod of length L and mass M and given by
cmL L / 2=
The net force acting on the rod is mg sin (forcealongedownwards)− θ
Torque is = force into perpendicular distance between the line of force and radius vecto
( ) mg sin( ) LLtorque( ) mg sin( )x 2 2
− θ 
∴ τ = − θ =   
4
The torque acting on the rod is
( ) mgsin( ) LLtorque( ) mg sin( )x 2 2
− θ 
τ = − θ =   
The angular acceleration is given by
p
(1)
I
Ï„
α = − − − − −
Where Ip = moment of inertia of the rod about the point P ( see in the diagram)
Ï„ = net torque acting on the rod

According to the parallel axis theorem
2
p cmI I MR= +
Where ICM = moment of inertia of the rod about the axis passing through the
centre of mass and parallel to the axis passing through P .

2
2
p
2
2 2
1 L
I ML M
12 2
1 L 1
ML M ML
12 4 3
 
= +  
 
= + =

substituting the values in the above equation , we get
2
mgsin L 3 3gsin
x
2 ML 2L
− θ θ
α = = −
(a)Lin Lear
L3gsin 3g sin
2L 2
momentum = α
− θ − θ
= =

At the initial position 90 sin 1θ = ∴ θ =�
23g(a) 1.5x9.8 14.7ms
2
Linear momentum −
−
= = − = −
Linear momentum = - 14.7 ms-2
5 At any instant , the bottom of this sphere is at rest on the surface. The axis passing through the point of
contact P parallel to its axis is called the instantaneous axis of rotation.
It has both translation and rotational motion, but it can be considered as pure rotation with an angular
velocity ω at that instant
2
p
1
KE of thesphere I
2
∴ = ω
Where Ip = moment of inertia of sphere about the instantaneous axis
ω = angular velocity
According to the parallel axis theorem
2
p cmI I MR= +
Where ICM = moment of inertia of sphere about the axis passing through the
centre of mass and parallel to the axis passing through P
2 2
cm
2 2 2
cm
1
KE (I MR )
2
1 1
I MR
2 2
∴ = + ω
= ω + ω
We are given that M =0.5 Kg , R= 10x10
-2
m

, v =12 ms
-1

1
-2
v 12
120 rad sec
R 10x10
−∴ ω = = =
2 2 2
cm
2 2 2 2
2 2 2 2
1
KE I MR
2
1 2
x MR MR
2 5
1 1 7
MR MR
5 2 10
∴ = ω + ω
= ω + ω
 
= + ω = ω 
 

after, substituting the numerical values in the above equation , we get
KE = 50.4 J

6. Let’s say that sphere will move to a height of h on an inclined plane ..
From energy conversation theorem we can say that the sphere will move till all its kinetic energy is
converted to potential energy
Hence, Mgh 50.4J=
Where M = Mass of the sphere
g = acceleration due to gravity (= 9.8ms
-2
)
50.4 50.4
h 10.3m
Mg 0.5x9.8
= = =
∴ h=10.3m
7.
Since there is no external torque acting on system about the rotation axis , the total angular momentum of the
system is conserved .
ie
( ) ( )
i,Tot f ,Tot
d d h h d h
2
d d d
2
h h h
L L
I I I I (1)
1 1
I MR x8x5x5x1.5 150 Nm (anticlosewise)
2 2
I MR 8x2.5x2.5x3 150 Nm (closewise)
=
∴ ω + ω = + ω − − − − − −
⇒ ω = ω = =
⇒ ω = ω = − = −
substituting the numerical values in the above equation(1) , we get
( ) ( )d hI I 150 150 0+ ω= − =
0∴ ω =
ie the final velocity of the combined system = ZERO

8
From the conservation of energy , we have
KE loss = total kinetic energy before collision - total kinetic energy of the coalition
( )
( )
disk hoop disk hoop
2 2 2
d d h h d h
2 2
d d h h
2
2
d
KE loss (KE KE ) (KE )
1 1 1
KE loss I I I I (where the variables have been defined in problem7)
2 2 2
But, the finalangular velocity 0
1 1
KE loss I I 0
2 2
1 MR 1
MR
2 2 2
+= + −
   
= ω + ω − + ω   
   
ω =
 
∴ = ω + ω − 
 
= ω + 2 2h
 
ω 
 

substituting the numerical values in the above equation , we get

2 2
2 21 8x5 8x2.5KE loss 1.5 3 33.75J
2 2 2
 
∴ = + = 
 

The loss of KE =338 J

9
There are two torques acting on the system one is the pτ exerted by the man which will rotate the system in anti-
clockwise direction and the other torque gτ exerted by the gravitational force...
SOLUTION.PDF
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here
Attach File