Question 1

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 4.2-rad displacement after starting from rest? A) 1.8 s B) 2.0 s C) 16 s D) 18 s

Accepted Answer

D

Question 2

A Ferris wheel,rotating initially at an angular speed of 0.50 rad/s,accelerates over a 7.0-s interval at a rate of 0.040 rad/s².What is its angular speed after this 7-s interval? A) 0.20 rad/s B) 0.30 rad/s C) 0.46 rad/s D) 0.78 rad/s

Accepted Answer

The final angular speed can be calculated using the formula: final speed = initial speed + (acceleration × time). So, 0.50 rad/s + (0.040 rad/s² × 7 s) = 0.78 rad/s.

Question 3

Starting from rest,a wheel undergoes constant angular acceleration for a period of time T.At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?
A) 0.25 T
B) 0.50 T
C) 0.67 T
D) 0.71 T

Accepted Answer

The average angular speed of the wheel during this interval is given by:
average angular speed = (final angular speed + initial angular speed) / 2
We know that the wheel starts from rest, so the initial angular speed is zero. Let the final angular speed be ωf. Using the kinematic equation:
ωf = ω0 + αt
where ω0 is the initial angular speed (zero in this case) and α is the angular acceleration, we can solve for the time t it takes to reach ωf:
t = (ωf - ω0) / α = ωf / α
So the wheel reaches an angular speed equal to its average angular speed when:
ωf = 2 * average angular speed
Substituting for ωf and solving for t, we get:
t = 2 * average angular speed / α = 0.5 * T
Therefore, the answer is B) 0.50 T.

Question 4

What is the angular speed about the rotational axis of the Earth for a person standing on the surface? A) 7.3 × 10−⁵ rad/s B) 3.6 × 10−⁵ rad/s C) 6.28 × 10−⁵ rad/s D) 3.14 × 10−⁵ rad/s

Accepted Answer

The Earth completes one rotation about its axis in approximately 24 hours (86400 seconds), so the angular speed is 2π radians/86400 seconds ≈ 7.3 × 10⁻⁵ rad/s.

Question 5

A 0.30-m-radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 2.0 rad/s² over a 5.0-s interval? A) 12.5 rad B) 25 rad C) 2.0 rad D) 0.50 rad

Accepted Answer

Using the equation 
θ = ½αt² + ω₀t
where θ is the angular displacement, α is the angular acceleration, ω₀ is the initial angular velocity and t is the time interval, we can calculate the total angular displacement of the tire:
θ = ½(2.0 rad/s²)(5.0 s)² + 0
θ = 50 rad
Therefore, the tire rotates 50 radians during the given time interval.

Question 6

A point on the rim of a 0.30-m-radius rotating wheel has a tangential speed of 4.0 m/s.What is the tangential speed of a point 0.20 m from the center of the same wheel?
A) 1.0 m/s
B) 1.3 m/s
C) 2.7 m/s
D) 8.0 m/s

Accepted Answer

The tangential speed is proportional to the radius for a point on a rotating wheel. If the speed at 0.30 m is 4.0 m/s, then at 0.20 m, it is (0.20/0.30) * 4.0 m/s = 2.7 m/s.

Question 7

A ceiling fan is turned on and reaches an angular speed of 120 rev/min in 20 s.It is then turned off and coasts to a stop in an additional 40 s.The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is which of the following?
A) 2
B) 0.5
C) −0.5
D) −2

Accepted Answer

The answer of A ceiling fan is turned on and...

Question 8

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 4.9 rad.What is its average angular acceleration? A) 0.10 rad/s² B) 0.05 rad/s² C) 1.8 rad/s² D) 0.60 rad/s²

Accepted Answer

The answer of A Ferris wheel starts at rest and...

Question 9

A spool of thread has an average radius of 1.00 cm.If the spool contains 62.8 m of thread,how many turns of thread are on the spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.
A) 100
B) 1 000
C) 3 140
D) 62 800

Accepted Answer

The answer of A spool of thread has an average...

Question 10

A ventilation fan has blades 0.25 m in radius rotating at 20 rpm.What is the tangential speed of each blade tip?
A) 0.02 m/s
B) 0.52 m/s
C) 5.0 m/s
D) 20 m/s

Accepted Answer

The tangential speed of each blade tip can be calculated using the formula:

v = rω

where v is the tangential speed, r is the radius and ω is the angular velocity.

Substituting the given values, we get:

v = 0.25 × 2π × 20

v = 0.25 × 40π

v ≈ 3.14 m/s

Therefore, the correct answer is B) 0.52 m/s (rounded to two decimal places).

Question 11

A 0.15-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) 6.8 m/s² C) 28 m/s² D) 14 m/s²

Accepted Answer

The average angular acceleration is (12.0 rad/s - 0 rad/s) / 4.0 s = 3.0 rad/s². The tangential acceleration is the angular acceleration multiplied by the radius: 3.0 rad/s² * 0.15 m = 0.45 m/s².

Question 12

2 600 rev/min is equivalent to which of the following?
A) 2600 rad/s
B) 43.3 rad/s
C) 273 rad/s
D) 60 rad/s

Accepted Answer

The answer of 2 600 rev/min is equivalent to which...

Question 13

Starting from rest,a wheel undergoes constant angular acceleration for a period of time T.At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?
A) 0.50 T
B) 0.67 T
C) 0.71 T
D) all of the above

Accepted Answer

The average angular acceleration is given by:
α_avg = Δω/Δt
where Δω is the change in angular velocity and Δt is the time interval over which the change occurs.

The instantaneous angular acceleration is given by:
α_inst = dω/dt
where dω is the instantaneous change in angular velocity and dt is the infinitesimal time interval over which the change occurs.

At any given time during the period T, the instantaneous angular acceleration is changing continuously, and so there is no one point in time where the average angular acceleration will exactly equal the instantaneous angular acceleration. However, as the time interval Δt becomes smaller and smaller, the average angular acceleration will approach the instantaneous angular acceleration. Therefore, at all times during the period T, the average angular acceleration will be approximately equal to the instantaneous angular acceleration, and so the correct answer is D, all of the above.

Question 14

A 0.30-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-s interval? A) 33 m/s² B) 6.7 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 = 0.30 \, \text{m} $. Thus, $ a_t = 2.0 \times 0.30 = 0.60 \, \text{m/s}^2 $.

Question 15

A grindstone spinning at the rate of 8.3 rev/s has what approximate angular speed?
A) 3.2 rad/s
B) 26 rad/s
C) 52 rad/s
D) 81 rad/s

Accepted Answer

To convert from revolutions per second to radians per second, we need to multiply by 2π, since there are 2π radians in one revolution. Thus, the angular speed of the grindstone is approximately 8.3 x 2π = 52 rad/s. Therefore, the answer is choice C.

Question 16

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 0.040 0 rad/s².What angular displacement does the Ferris wheel undergo in this 7-s interval? A) 4.48 rad B) 2.50 rad C) 3.00 rad D) 0.500 rad

Accepted Answer

The angular displacement can be calculated using the formula: θ = ω_initial * t + 0.5 * α * t^2. Substituting the given values: θ = 0.500 * 7 + 0.5 * 0.040 * 7^2 = 4.48 rad.

Question 17

A 0.12-m-radius grinding wheel takes 5.5 s to speed up from 2.0 rad/s to 11.0 rad/s.What is the wheel's average angular acceleration? A) 9.6 rad/s² B) 4.8 rad/s² C) 1.6 rad/s² D) 0.33 rad/s²

Accepted Answer

The answer of A 0.12-m-radius grinding wheel takes 5.5 s...

Question 18

A ceiling fan is turned on and reaches an angular speed of 120 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) 20
B) 60
C) 0
D) 600

Accepted Answer

In the first 20 seconds, the fan rotates at a constant angular speed of 120 rev/min. 
Therefore, the number of revolutions in 20 seconds = (120 rev/min)(20/60 min) = 40 revolutions.

In the next 40 seconds, the fan coasts to a stop. 
Since the fan was rotating initially, its final angular speed is zero. 
Using the formula, final angular speed = initial angular speed + angular acceleration × time, 
we can find the angular acceleration, which is constant during the coasting period: 
0 = 2π/60 × (120/60) + angular acceleration × 40 
Solving for angular acceleration, we get: 
angular acceleration = -π/240 rad/s^2

Using the formula, θ = (1/2) × angular acceleration × time^2, 
we can find the total number of revolutions during the coasting period: 
θ = (1/2) × (-π/240) × (40^2) = -6.67 revolutions (the negative sign indicates that the fan rotated in the opposite direction during coasting)

Therefore, the total number of revolutions in one minute = 40 + (-6.67) = 33.33 revolutions, which is closest to 60 revolutions (option B) when rounded to the nearest integer.

Question 19

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 4.2 rad? A) 0.025 rad/s B) 0.11 rad/s C) 0.46 rad/s D) 1.2 rad/s

Accepted Answer

Using the angular kinematic equation, 
θ = ω_i*t + (1/2)*α*t^2

where θ is the angular displacement, ω_i is the initial angular speed (which is zero), α is the constant angular acceleration, and t is time.

We want to find ω_f, the final angular speed when θ = 4.2 rad. Rearranging the equation and plugging in the given values,

ω_f = sqrt(2*α*θ) 
= sqrt(2*0.025 rad/s^2 * 4.2 rad) 
= 0.46 rad/s

Therefore, the answer is C) 0.46 rad/s.

Question 20

Suppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s after 20.0 seconds have elapsed.How long after the initial time has the wheel undergone half of the angular displacement that it will have gone through during the entire 20.0 second interval?
A) 10.0 s
B) 12.4 s
C) 14.2 s
D) 15.0 s

Accepted Answer

The answer of Suppose a wheel is initially rotating at...

Quiz 7: Rotational Motion and the Law of Gravity

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 4.2-rad displacement after starting from rest?

A Ferris wheel,rotating initially at an angular speed of 0.50 rad/s,accelerates over a 7.0-s interval at a rate of 0.040 rad/s2.What is its angular speed after this 7-s interval?

Starting from rest,a wheel undergoes constant angular acceleration for a period of time T.At what time after the start of rotation does the wheel reach an angular speed equal to its average angular speed for this interval?

What is the angular speed about the rotational axis of the Earth for a person standing on the surface?

A 0.30-m-radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 2.0 rad/s2 over a 5.0-s interval?

A point on the rim of a 0.30-m-radius rotating wheel has a tangential speed of 4.0 m/s.What is the tangential speed of a point 0.20 m from the center of the same wheel?

A ceiling fan is turned on and reaches an angular speed of 120 rev/min in 20 s.It is then turned off and coasts to a stop in an additional 40 s.The ratio of the average angular acceleration for the first 20 s to that for the last 40 s is which of the following?

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 4.9 rad.What is its average angular acceleration?

A spool of thread has an average radius of 1.00 cm.If the spool contains 62.8 m of thread,how many turns of thread are on the spool? "Average radius" allows us to not need to treat the layering of threads on lower layers.

A ventilation fan has blades 0.25 m in radius rotating at 20 rpm.What is the tangential speed of each blade tip?

A 0.15-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?

2 600 rev/min is equivalent to which of the following?

Starting from rest,a wheel undergoes constant angular acceleration for a period of time T.At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?

A 0.30-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-s interval?

A grindstone spinning at the rate of 8.3 rev/s has what approximate angular speed?

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 0.040 0 rad/s2.What angular displacement does the Ferris wheel undergo in this 7-s interval?

A 0.12-m-radius grinding wheel takes 5.5 s to speed up from 2.0 rad/s to 11.0 rad/s.What is the wheel's average angular acceleration?

A ceiling fan is turned on and reaches an angular speed of 120 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 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 4.2 rad?

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 4.2-rad displacement after starting from rest?

A Ferris wheel,rotating initially at an angular speed of 0.50 rad/s,accelerates over a 7.0-s interval at a rate of 0.040 rad/s².What is its angular speed after this 7-s interval?

A 0.30-m-radius automobile tire rotates how many rad after starting from rest and accelerating at a constant 2.0 rad/s² over a 5.0-s interval?

A 0.30-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-s interval?

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 0.040 0 rad/s².What angular displacement does the Ferris wheel undergo in this 7-s interval?

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 4.2 rad?