Question 1

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. -What is the magnitude of the centripetal acceleration of a point on the rim of the grindstone? A)zero m/s² B)0.5 m/s² C)1.0 m/s² D)2.0 m/s² E)4.0 m/s²

Accepted Answer

The centripetal acceleration is given by the formula $ a_c = \omega^2 \times r $, where $ \omega = 2.0 \, \text{rad/s} $ and $ r = 1.0 \, \text{m} $. Thus, $ a_c = (2.0)^2 \times 1.0 = 4.0 \, \text{m/s}^2 $.

Question 2

A top is spinning counterclockwise as shown in the figure. It is also moving to the right with a linear speed v. What is the direction of the angular velocity? 
A)upward
B)downward
C)left
D)right
E)into the paper

Accepted Answer

The answer of A top is spinning counterclockwise as shown...

Question 3

A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.  
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the direction of the disk's angular velocity?
A)to the left
B)to the right
C)into the paper (away from you)
D)out of the paper (toward you)
E)It varies from point to point on the disk.

Accepted Answer

The direction of angular velocity is determined by the right-hand rule. For a disk rotating in a plane, if you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the angular velocity vector, which is typically into the paper (away from you) for counterclockwise rotation.

Question 4

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center.
-What is the ratio of the centripetal acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod?
A)1:1
B)1:2
C)2:1
D)4:1
E)1:4

Accepted Answer

The centripetal acceleration is given by:

a_c = rω^2

where r is the distance from the axis of rotation and ω is the angular velocity.

At the end of the rod, the distance from the axis of rotation is L and the angular velocity is ω = αt, where α is the angular acceleration and t is the time.

Therefore, the centripetal acceleration at the end of the rod is:

a_c1 = L(αt)^2

At a distance of L/2 from the end of the rod, the distance from the axis of rotation is L/2 and the angular velocity is half that of the end of the rod: ω = 1/2 αt.

Therefore, the centripetal acceleration at a distance of L/2 from the end of the rod is:

a_c2 = (L/2)(1/2 αt)^2 = (L/8)α^2t^2

Taking the ratio of these two accelerations, we get:

a_c1/a_c2 = [L(αt)^2]/[(L/8)α^2t^2] = 8

So the ratio of the centripetal acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod is 2:1.

Question 5

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center.
-What is the ratio of the tangential acceleration of a point on the end of the rod to that of a point a distance L/2 from the end of the rod?
A)1:1
B)1:2
C)2:1
D)4:1
E)1:4

Accepted Answer

The tangential acceleration is given by $ a_t = \alpha \times r $, where $ \alpha $ is the angular acceleration and $ r $ is the distance from the axis of rotation. For a point at the end of the rod, $ r = L $, and for a point at $ L/2 $ from the end, $ r = L/2 $. Thus, the ratio is $ L : (L/2) = 2:1 $.

Question 6

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s.
-How far did the bicycle travel during the 5.0 seconds of braking?
A)1.8 m
B)8.8 m
C)22 m
D)42 m
E)44 m

Accepted Answer

First, we need to find the initial linear speed of the bicycle. Since it is rolling without slipping, the linear speed is equal to the product of the angular speed and the radius:
v = ωr = (2.0 rev/s)(2π)(0.70 m) = 8.8 m/s

During the 5.0 s of braking, the bicycle experiences a constant deceleration that causes it to stop. The deceleration can be found using the kinematic equation:
Δx = v0t + ½at^2

where Δx is the distance traveled during the braking, v0 is the initial speed, a is the deceleration, and t is the time.

Since the bicycle stops, its final speed is zero, so we have:
0 = v0 - at
a = v0/t = 8.8 m/s / 5.0 s = 1.8 m/s^2

Substituting this into the kinematic equation, we get:
Δx = v0t + ½at^2 = (8.8 m/s)(5.0 s) + ½(1.8 m/s^2)(5.0 s)^2 = 22 m

Therefore, the bicycle traveled a distance of 22 m during the 5.0 s of braking. Choice C is the best answer.

Question 7

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center.
-What is the ratio of the angular speed (at any instant) of a point on the end of the rod to that of a point a distance L/2 from the end of the rod?
A)1:1
B)1:2
C)2:1
D)4:1
E)1:4

Accepted Answer

The angular speed is the same for all points on the rod because they all rotate together as a rigid body. Therefore, the ratio of the angular speed of a point on the end of the rod to that of a point a distance L/2 from the end is 1:1.

Question 8

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. -What is the magnitude of the tangential acceleration of a point on the rim of the wheel? A)zero m/s² B)0.5 m/s² C)1.0 m/s² D)2.0 m/s² E)4.0 m/s²

Accepted Answer

The tangential acceleration is zero because the grindstone is spinning with a constant angular speed, meaning there is no change in the angular velocity.

Question 9

A car is moving along a horizontal road at a constant velocity that is directed 45° south of east. What is the direction of the angular velocity of the wheels of the car?
A)45° south of west
B)45° north of west
C)45° south of east
D)45° north of east
E)due east

Accepted Answer

The direction of the angular velocity of the wheels of the car is perpendicular to the direction of motion and in the direction of the rotation. Since the car is moving 45° south of east, the direction of its velocity is also 45° north of east. Therefore, the direction of the angular velocity of the wheels must be 45° perpendicular to this, which is 45° north of east. Choice D is the only option that matches this direction.

Question 10

A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.  
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the instantaneous speed of point A with respect to the ground?
A)zero m/s
B)2.5 m/s
C)5.0 m/s
D)7.1 m/s
E)10.0 m/s

Accepted Answer

The answer of A 2.0-kg solid disk rolls without slipping...

Question 11

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s.
-Through how many revolutions did the wheel rotate during the 5.0 seconds of braking?
A)10 rev
B)2.0 rev
C)9.6 rev
D)5.0 rev
E)0.4 rev

Accepted Answer

The answer of A bicycle wheel of radius 0.70 m...

Question 12

A long thin rod of length 2L rotates with a constant angular acceleration of 8.0 rad/s2 about an axis that is perpendicular to the rod and passes through its center.
-What is the ratio of the tangential speed (at any instant) of a point on the end of the rod to that of a point a distance L/2 from the end of the rod?
A)1:1
B)1:2
C)2:1
D)4:1
E)1:4

Accepted Answer

The tangential acceleration of a point on the end of the rod can be found using the equation:

a_t = r * alpha

where a_t is the tangential acceleration, r is the distance from the axis of rotation to the point, and alpha is the angular acceleration. At the end of the rod, r = L, so:

a_t1 = L * 8.0 = 8L

Similarly, for a point a distance L/2 from the end, r = 3L/2, so:

a_t2 = (3L/2) * 8.0 = 12L

The tangential velocity at any instant can be found using the equation:

v_t = r * omega

where v_t is the tangential velocity, r is the distance from the axis of rotation to the point, and omega is the angular velocity. Assuming that both points start from rest, the ratio of their tangential speeds will be the same as the ratio of their tangential accelerations:

v_t1/v_t2 = a_t1/a_t2 = (8L)/(12L) = 2/3

Therefore, the ratio of the tangential speeds of the two points is 2:3, or equivalently, 2:1. The closest choice is C) 2:1.

Question 13

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s.
-What is the tangential speed of a point on the rim of the grindstone?
A)zero m/s
B)0.5 m/s
C)1.0 m/s
D)2.0 m/s
E)4.0 m/s

Accepted Answer

The tangential speed of a point on the rim of the grindstone is given by:
v = rω
where v is tangential speed, r is the radius of the grindstone, and ω is the angular speed.
Plugging in the given values, we get:
v = (1.0 m)(2.0 rad/s) = 2.0 m/s
Therefore, the answer is (D) 2.0 m/s.

Question 14

A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.  
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the instantaneous speed of the point of the disk that makes contact with the surface?
A)zero m/s
B)5.0 m/s
C)7.1 m/s
D)7.5 m/s
E)10.0 m/s

Accepted Answer

The instantaneous speed of the point of the disk that makes contact with the surface is zero because, at that instant, the point is momentarily at rest relative to the surface.

Question 15

A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.  
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the instantaneous speed of point B with respect to the ground?
A)zero m/s
B)5.0 m/s
C)7.1 m/s
D)7.5 m/s
E)10.0 m/s

Accepted Answer

The instantaneous speed of point B can be found using the equation v = ωr, where ω is the angular velocity and r is the distance from the center of rotation to point B. Since the disk is rolling without slipping, we know that the linear velocity of the center of the disk is v = 5.0 m/s, and the angular velocity is ω = v/r.

To find r, we can use the fact that the center of the disk and point B lie on a horizontal line, so the distance between them is the radius of the disk. The radius of the disk is not provided, but we can calculate it using the moment of inertia of a solid disk about its center, I = (1/2)mr^2.

Solving for r, we have r = sqrt(2I/m) = sqrt(1/2) * sqrt(I/m) = (1/sqrt(2)) * sqrt((1/2) * (2.0 kg) * (0.5 m)^2) = 0.5 m.

Thus, ω = v/r = (5.0 m/s) / (0.5 m) = 10.0 rad/s, and vB = ωr = (10.0 rad/s) * (0.5 m) = 5.0 m/s. Therefore, the instantaneous speed of point B with respect to the ground is 5.0 m/s.

Quiz 8: Rotational Kinematics

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. -What is the magnitude of the centripetal acceleration of a point on the rim of the grindstone?

A top is spinning counterclockwise as shown in the figure. It is also moving to the right with a linear speed v. What is the direction of the angular velocity?

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s. -How far did the bicycle travel during the 5.0 seconds of braking?

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. -What is the magnitude of the tangential acceleration of a point on the rim of the wheel?

A car is moving along a horizontal road at a constant velocity that is directed 45° south of east. What is the direction of the angular velocity of the wheels of the car?

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. The bicycle stops in 5.0 s. -Through how many revolutions did the wheel rotate during the 5.0 seconds of braking?

A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. -What is the tangential speed of a point on the rim of the grindstone?