Question 1

A bucket of water with total mass 30 kg is attached to a rope,which in turn is wound around a 0.050 m radius cylinder at the top of a well.A crank with a turning radius of 0.25 m is attached to the end of the cylinder and the moment of inertia of cylinder and crank is 0.12 kg×m².If the bucket is raised to the top of the well and released,what is the acceleration of the bucket as it falls toward the bottom of the well? (Assume rope's mass is negligible,that cylinder turns on frictionless bearings and that g = 9.8 m/s². ) A) 3.2 m/s² B) 3.8 m/s² C) 7.4 m/s² D) 9.8 m/s²

Accepted Answer

The answer of A bucket of water with total mass...

Question 2

If a nonzero torque is applied to an object,that object will experience:
A) a constant angular speed.
B) an angular acceleration.
C) a constant moment of inertia.
D) an increasing moment of inertia.

Accepted Answer

When a nonzero torque is applied to an object, it will experience an angular acceleration. This is because torque is directly proportional to the angular acceleration of an object, according to Newton's Second Law for Rotation. The moment of inertia, which is the rotational equivalent of mass, remains constant unless the object's shape or distribution of mass changes. Therefore, choices A, C, and D are incorrect.

Question 3

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s².What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s? A) 4.0 m/s² B) 40 m/s² C) 44 m/s² D) 400 m/s²

Accepted Answer

The answer of A 0.40-m radius automobile tire accelerates from...

Question 4

An automobile accelerates from zero to 30 m/s in 6.0 s.The wheels have a diameter of 1.0 m.What is the average angular acceleration of each wheel? A) 5.0 rad/s² B) 10 rad/s² C) 25 rad/s² D) 40 rad/s²

Accepted Answer

The linear acceleration is 30 m/s divided by 6.0 s, which is 5 m/s². The radius of the wheel is 0.5 m (half of the diameter). The angular acceleration is the linear acceleration divided by the radius, which is 5 m/s² / 0.5 m = 10 rad/s².

Question 5

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s².What is its angular speed at the instant it goes through an angular displacement of 10 rad? A) 0.025 rad/s B) 0.11 rad/s C) 0.46 rad/s D) 0.71 rad/s

Accepted Answer

Using the equation ω² = ω₀² + 2αθ, where ω₀ = 0, α = 0.025 rad/s², and θ = 10 rad, we find ω = √(2 * 0.025 * 10) = 0.71 rad/s.

Question 6

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 2.5 rad.What is its average angular acceleration? A) 0.098 rad/s² B) 0.05 rad/s² C) 1.8 rad/s² D) 0.60 rad/s²

Accepted Answer

We can use the formula for average angular acceleration:

average angular acceleration = change in angular speed / change in time

The final angular speed is given as 0.70 rad/s and the initial angular speed is 0 (since the Ferris wheel starts at rest). The time it takes to rotate through an angular displacement of 2.5 rad can be found using the formula:

angular displacement = (average angular speed)(time)

rearranging this formula gives:

time = angular displacement / average angular speed

Plugging in the values we get:

time = 2.5 rad / 0.70 rad/s = 3.57 s

Therefore, the change in angular speed is:

change in angular speed = final angular speed - initial angular speed = 0.70 rad/s - 0 rad/s = 0.70 rad/s

So, the average angular acceleration is:

average angular acceleration = change in angular speed / change in time = 0.70 rad/s / 3.57 s = 0.098 rad/s^2

The closest answer choice to this value is A: 0.098 rad/s^2, so that is the correct answer.

Question 7

A rod of length L is hinged at one end.The moment of inertia as the rod rotates around that hinge is ML²/3.Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position.The rod is released as the free end is allowed to fall.What is the angular acceleration as it is released? A) 3.70 rad/s² B) 7.35 rad/s² C) 5.88 rad/s² D) 4.90 rad/s²

Accepted Answer

The moment of inertia (I) and the mass (m) of the rod are given. To find the angular acceleration (α), we can use the equation: τ = Iα where τ is the torque applied to the rod. In this case, the only torque acting on the rod is the torque due to gravity: τ = mgl/2 where g is the acceleration due to gravity and l is the length of the rod. Therefore, we can write: mgl/2 = Iα Solving for α, we get: α = mgl / 2I Substituting the given values, we get: α = (3.00 kg)(9.81 m/s^2)(2.50 m) / [2(ML²/3)] α ≈ 5.88 rad/s^2 Therefore, the answer is C, 5.88 rad/s^2.

Question 8

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.050 m.How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.20 m/s?
A) 3.0 rev/s
B) 1.5 rev/s
C) 0.48 rev/s
D) 0.64 rev/s

Accepted Answer

The answer of A bucket in an old well is...

Question 9

A ventilation fan has blades 0.25 m in radius rotating at 30 rpm.What is the tangential speed of each blade tip?
A) 0.02 m/s
B) 0.52 m/s
C) 0.79 m/s
D) 20 m/s

Accepted Answer

Given, 
Radius of the blade, r = 0.25 m
Rotational speed, ω = 30 rpm
Tangential speed of the blade tip, V = rω
Substituting the values, we get
V = 0.25 * 2π * 30 m/s = 0.79 m/s
Therefore, the tangential speed of each blade tip is 0.79 m/s, which is given in option C.

Question 10

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s² over a 5.0-s interval? A) 12 rad B) 25 rad C) 37 rad D) 0.50 rad

Accepted Answer

The initial angular velocity is zero, and the constant angular acceleration is 3.0 rad/s^2. Using the equation:

θ = ω_i*t + (1/2)*α*t^2

where θ is the total angular displacement, ω_i is the initial angular velocity, α is the angular acceleration, and t is the time interval, we can calculate the total angular displacement:

θ = (0)*5.0 + (1/2)*(3.0)*(5.0)^2 = 37.5 rad

However, this is the total displacement, so we need to subtract the initial angle (which is zero) to get the net angular displacement:

θ_net = 37.5 rad - 0 rad = 37.5 rad

Therefore, the tire rotates 37.5 radians during the 5.0-second interval, which is closest to choice C, 37 rad.

Question 11

A ventilation fan with a moment of inertia of 0.034 kg·m² has a net torque of 0.14 N·m applied to it.What angular acceleration does it experience? A) 5.3 rad/s² B) 4.1 rad/s² C) 3.2 rad/s² D) 0.31 rad/s²

Accepted Answer

The answer of A ventilation fan with a moment of...

Question 12

A Ferris wheel,rotating initially at an angular speed of 0.500 rad/s,accelerates over a 7.00-s interval at a rate of 0.0500 rad/s².What angular displacement does the Ferris wheel undergo in this 7.00-s interval? A) 4.48 rad B) 2.50 rad C) 4.73 rad D) 0.500 rad

Accepted Answer

We can use the formula

θ = ωi*t + (1/2) * α * t^2

where 
ωi = 0.500 rad/s (initial angular speed) 
α = 0.0500 rad/s^2 (angular acceleration) 
t = 7.00 s (time interval)

Plugging in the values, we get

θ = (0.500 rad/s)*(7.00 s) + (1/2)*(0.0500 rad/s^2)*(7.00 s)^2 
θ ≈ 4.73 rad

Therefore, the Ferris wheel undergoes an angular displacement of approximately 4.73 radians during this 7.00-second interval. The correct answer is (C).

Question 13

A 0.20-m radius grinding wheel starts at rest and develops an angular speed of 12.0 rad/s in 4.0 s.What is the average tangential acceleration of a point on the wheel's edge? A) 0.45 m/s² B) 0.60 m/s² C) 28 m/s² D) 14 m/s²

Accepted Answer

We can use the formula for angular acceleration:

α = Δω/Δt

where α is the angular acceleration, Δω is the change in angular speed and Δt is the time taken.

Rearranging, we get:

Δω = αΔt

The tangential acceleration can be found using the formula:

at = rα

where r is the radius of the wheel.

So, we can substitute the given values and calculate:

Δω = 12.0 rad/s - 0 rad/s = 12.0 rad/s

Δt = 4.0 s

α = Δω/Δt = 3.0 rad/s^2

at = rα = 0.20 m × 3.0 rad/s^2 = 0.60 m/s^2

Therefore, the average tangential acceleration of a point on the wheel's edge is 0.60 m/s^2, which corresponds to choice B.

Question 14

A disk has a moment of inertia of 3.0 × 10−⁴ kg·m² and rotates with an angular speed of 3.5 rad/sec.What net torque must be applied to bring it to rest within 3.7 s? A) 4.6 × 10−⁴ N·m B) 7.5 × 10−⁴ N·m C) 3.5 × 10−⁴ N·m D) 2.8 × 10−⁴ N·m

Accepted Answer

The answer of A disk has a moment of inertia...

Question 15

A Ferris wheel,rotating initially at an angular speed of 0.16 rad/s,accelerates over a 7.0-s interval at a rate of 0.050 rad/s².What is its angular speed after this 7.0-s interval? A) 0.46 rad/s B) 0.85 rad/s C) 0.51 rad/s D) 1.2 rad/s

Accepted Answer

Using the formula:

ω = ω0 + αt

where 
ω = final angular speed 
ω0 = initial angular speed = 0.16 rad/s 
α = angular acceleration = 0.050 rad/s² 
t = time interval = 7.0 s

Substituting the values:

ω = 0.16 + (0.050)(7.0) 
ω = 0.51 rad/s

Therefore, the final angular speed of the Ferris wheel after the 7.0-s interval is 0.51 rad/s. 
The correct choice is C.

Question 16

A baseball pitcher,loosening up his arm before a game,tosses a 0.15-kg ball using only the rotation of his forearm,0.32 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 24 m/s in a time of 0.40 s,what is the average angular acceleration of the arm and ball? A) 0.067 rad/s² B) 94 rad/s² C) 15 rad/s² D) 190 rad/s²

Accepted Answer

The answer of A baseball pitcher,loosening up his arm before...

Question 17

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s².What is the time interval required for it to reach a 10-rad displacement after starting from rest? A) 28 s B) 2.0 s C) 56 s D) 18 s

Accepted Answer

Explanation:Using the equation of motion,$$	heta_f=	heta_i+\omega_it+\frac{1}{2}\alpha t^2$$$$10=0+0+\frac{1}{2}	imes0.025	imes t^2$$$$t^2=800$$$$t=28.28\approx28	ext{ s}$$

Question 18

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 20 s.It is then turned off and coasts to a stop in 40 s.In the one minute of rotation,through how many revolutions did the fan turn?
A) 30
B) 60
C) 0
D) 600

Accepted Answer

The fan accelerates to 60 rev/min in 20 s, which is 1/3 of a minute. The average speed during this time is 30 rev/min, so it completes 10 revolutions. After being turned off, it coasts to a stop in 40 s (2/3 of a minute) with an average speed of 30 rev/min, completing 20 revolutions. Total revolutions = 10 + 20 = 30.

Question 19

A baseball pitcher loosens up his pitching arm.He tosses a 0.15-kg ball using only the rotation of his forearm,0.32 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 24 m/s in a time of 0.40 s,what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?
A) 1.1 N·m
B) 11 N·m
C) 2.9 N·m
D) 1.4 N·m

Accepted Answer

We can use the formula for torque:

τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

We can use the formula for moment of inertia for a point mass:

I = mr^2

Where m is the mass of the ball and r is the length of the pitcher's forearm.

Plugging in the given values:

I = (0.15 kg)(0.32 m)^2 = 0.01536 kg·m^2

We can rearrange the formula for torque to solve for it:

τ = Iα

τ = (0.01536 kg·m^2)(24 m/s / 0.40 s)

τ = 0.9216 N·m

Therefore, the torque applied to the ball is 0.9216 N·m. The closest answer choice to this value is C) 2.9 N·m.

Question 20

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval? A) 33 m/s² B) 0.80 m/s² C) 0.60 m/s² D) 0.30 m/s²

Accepted Answer

The tangential acceleration is given by the formula $ a_t = \alpha \times r $, where $ \alpha = 2.0 \, \text{rad/s}^2 $ and $ r = 0.40 \, \text{m} $. Thus, $ a_t = 2.0 \times 0.40 = 0.80 \, \text{m/s}^2 $.

Quiz 8: Rotational Equilibrium and Rotational Dynamics: Part A

If a nonzero torque is applied to an object,that object will experience:

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s2.What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s?

An automobile accelerates from zero to 30 m/s in 6.0 s.The wheels have a diameter of 1.0 m.What is the average angular acceleration of each wheel?

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s2.What is its angular speed at the instant it goes through an angular displacement of 10 rad?

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 2.5 rad.What is its average angular acceleration?

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.050 m.How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.20 m/s?

A ventilation fan has blades 0.25 m in radius rotating at 30 rpm.What is the tangential speed of each blade tip?

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s2 over a 5.0-s interval?

A ventilation fan with a moment of inertia of 0.034 kg·m2 has a net torque of 0.14 N·m applied to it.What angular acceleration does it experience?

A Ferris wheel,rotating initially at an angular speed of 0.500 rad/s,accelerates over a 7.00-s interval at a rate of 0.0500 rad/s2.What angular displacement does the Ferris wheel undergo in this 7.00-s interval?

A 0.20-m radius grinding wheel starts at rest and develops an angular speed of 12.0 rad/s in 4.0 s.What is the average tangential acceleration of a point on the wheel's edge?

A disk has a moment of inertia of 3.0 × 10−4 kg·m2 and rotates with an angular speed of 3.5 rad/sec.What net torque must be applied to bring it to rest within 3.7 s?

A Ferris wheel,rotating initially at an angular speed of 0.16 rad/s,accelerates over a 7.0-s interval at a rate of 0.050 rad/s2.What is its angular speed after this 7.0-s interval?

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s2.What is the time interval required for it to reach a 10-rad displacement after starting from rest?

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 20 s.It is then turned off and coasts to a stop in 40 s.In the one minute of rotation,through how many revolutions did the fan turn?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s2 over a 5.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s².What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s?

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s².What is its angular speed at the instant it goes through an angular displacement of 10 rad?

A 0.30-m radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad/s² over a 5.0-s interval?

A ventilation fan with a moment of inertia of 0.034 kg·m² has a net torque of 0.14 N·m applied to it.What angular acceleration does it experience?

A Ferris wheel,rotating initially at an angular speed of 0.500 rad/s,accelerates over a 7.00-s interval at a rate of 0.0500 rad/s².What angular displacement does the Ferris wheel undergo in this 7.00-s interval?

A disk has a moment of inertia of 3.0 × 10−⁴ kg·m² and rotates with an angular speed of 3.5 rad/sec.What net torque must be applied to bring it to rest within 3.7 s?

A Ferris wheel,rotating initially at an angular speed of 0.16 rad/s,accelerates over a 7.0-s interval at a rate of 0.050 rad/s².What is its angular speed after this 7.0-s interval?

A fan blade,initially at rest,rotates with a constant acceleration of 0.025 rad/s².What is the time interval required for it to reach a 10-rad displacement after starting from rest?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval?