additional practice
1. what is the angular displacement of each of the following hands of a clock in 1.00 h? state your answer in three significant digits.
a. the second hand
b. the minute hand
c. the hour hand
2. a rotating toy above a crib makes one complete counterclockwise rotation in 1 min.
a. what is its angular displacement in 3 min?
b. what is the toy’s angular velocity in rad/min?
c. if the toy is turned off, does it have positive or negative angular acceleration? explain.
3. if a truck has a linear acceleration of 1.85 m/s² and the wheels have an angular acceleration of 5.23 rad/s², what is the diameter of the wheels?
4. the truck in the previous problem is towing a trailer with wheels that have a diameter of 48 cm.
a. how does the linear acceleration of the trailer compare with that of the truck?
b. how do the angular accelerations of the wheels of the trailer and the wheels of the truck compare?
5. challenge you replace the tires on your car with tires of larger diameter. how will the angular velocity and number of revolutions change for trips at the same speed and over the same distance?

Question

additional practice
1. what is the angular displacement of each of the following hands of a clock in 1.00 h? state your answer in three significant digits.
   a. the second hand
   b. the minute hand
   c. the hour hand
2. a rotating toy above a crib makes one complete counterclockwise rotation in 1 min.
   a. what is its angular displacement in 3 min?
   b. what is the toy’s angular velocity in rad/min?
   c. if the toy is turned off, does it have positive or negative angular acceleration? explain.
3. if a truck has a linear acceleration of 1.85 m/s² and the wheels have an angular acceleration of 5.23 rad/s², what is the diameter of the wheels?
4. the truck in the previous problem is towing a trailer with wheels that have a diameter of 48 cm.
   a. how does the linear acceleration of the trailer compare with that of the truck?
   b. how do the angular accelerations of the wheels of the trailer and the wheels of the truck compare?
5. challenge you replace the tires on your car with tires of larger diameter. how will the angular velocity and number of revolutions change for trips at the same speed and over the same distance?

Accepted Answer

### Question 1:
#### Part a: Second Hand
# Explanation:
## Step 1: Determine the time and full rotation angle
The second hand completes a full rotation ($2\pi$ radians) in 60 seconds. In 1.00 hour, there are $1.00 	imes 3600 = 3600$ seconds.
## Step 2: Calculate the number of full rotations
Number of rotations $n = \frac{3600}{60} = 60$.
## Step 3: Calculate the angular displacement
Angular displacement $	heta = n 	imes 2\pi = 60 	imes 2\pi = 120\pi \approx 377$ radians. But since it's a clock, the angular displacement is negative (clockwise) or we can consider the net displacement. Wait, actually, for a clock, the second hand moves clockwise. In 1 hour, it makes 60 revolutions. The angular displacement is $- 60	imes 2\pi=- 120\pi\approx - 377$ radians (negative because clockwise). But if we consider the magnitude for the question (maybe they want the magnitude or the actual displacement). Wait, the question says "angular displacement". Let's re - check. The second hand in 1 hour: each minute it does 60 seconds, so in 60 minutes (1 hour), it does 60*60 = 3600 seconds. Each second it moves $2\pi/60$ radians. So total angular displacement $	heta=\omega t$, where $\omega = - 2\pi/60$ rad/s (negative for clockwise), $t = 3600$ s. So $	heta=-\frac{2\pi}{60}	imes3600=- 120\pi\approx - 377$ rad. Rounding to three significant digits, it's $- 3.77	imes 10^{2}$ rad or $3.77	imes 10^{2}$ rad (if we consider magnitude, but direction matters for angular displacement). But usually, for clock hands, clockwise is negative.
## Step 4: Final calculation
$	heta=- 120\pi\approx - 377$ rad (or $3.77	imes 10^{2}$ rad with three significant digits).

#### Part b: Minute Hand
# Explanation:
## Step 1: Time and full rotation angle
The minute hand completes a full rotation ($2\pi$ radians) in 60 minutes. In 1.00 hour (60 minutes), it completes 1 full rotation.
## Step 2: Angular displacement
Angular displacement $	heta=- 2\pi\approx - 6.28$ radians (negative because clockwise).

#### Part c: Hour Hand
# Explanation:
## Step 1: Time and full rotation angle
The hour hand completes a full rotation ($2\pi$ radians) in 12 hours. In 1.00 hour, the fraction of the rotation is $\frac{1}{12}$.
## Step 2: Angular displacement
Angular displacement $	heta =-\frac{1}{12}	imes2\pi=-\frac{\pi}{6}\approx - 0.524$ radians (negative because clockwise).

### Question 2:
#### Part a: Angular Displacement in 3 min
# Explanation:
## Step 1: Determine the angular velocity
The toy makes 1 complete counter - clockwise rotation in 1 min. So the angular velocity $\omega=\frac{2\pi}{1\ min}=2\pi$ rad/min.
## Step 2: Calculate angular displacement
Using $	heta=\omega t$, where $t = 3$ min. So $	heta=2\pi	imes3 = 6\pi\approx18.8$ radians.

#### Part b: Angular Velocity in rad/min
# Explanation:
The toy makes 1 full rotation ($2\pi$ radians) in 1 minute. So angular velocity $\omega=\frac{2\pi}{1\ min}=2\pi\approx6.28$ rad/min.

#### Part c: Angular Acceleration when turned off
# Explanation:
When the toy is turned off, it will slow down (assuming friction or some resistance). Since it was rotating counter - clockwise, and now it is decelerating (slowing down), the angular acceleration will be clockwise. In the convention where counter - clockwise is positive, the angular acceleration will be negative. Because the angular velocity (positive, counter - clockwise) is decreasing, so the angular acceleration (which is the rate of change of angular velocity) will be in the opposite direction of the angular velocity, i.e., negative.

### Question 3:
# Explanation:
## Step 1: Recall the relationship between linear and angular acceleration
The relationship between linear acceleration $a$ and angular acceleration $\alpha$ is $a = r\alpha$, where $r$ is the radius of the wheel. The radius $r=\frac{d}{2}$, where $d$ is the diameter.
## Step 2: Rearrange the formula to solve for $d$
We know that $a = r\alpha=\frac{d}{2}\alpha$, so $d=\frac{2a}{\alpha}$.
## Step 3: Substitute the given values
Given $a = 1.85$ m/s² and $\alpha = 5.23$ rad/s².
$d=\frac{2	imes1.85}{5.23}=\frac{3.7}{5.23}\approx0.707$ m. So the diameter is approximately $0.707$ m or 70.7 cm.

### Question 4:
#### Part a: Linear Acceleration Comparison
# Explanation:
The linear acceleration of a point on the wheel is related to the angular acceleration by $a = r\alpha$. For the truck and the trailer, since they are towed together, the linear acceleration of the trailer (the linear acceleration of the wheels of the trailer) should be equal to the linear acceleration of the truck. Because the truck is towing the trailer, they move with the same linear acceleration (assuming no slipping and the connection is rigid). The angular acceleration of the wheels of the trailer and the truck may be different if their wheel diameters are different, but the linear acceleration of the trailer (the linear acceleration of its wheels' edge) should be equal to that of the truck.

#### Part b: Angular Acceleration Comparison
# Explanation:
We know that $a = r\alpha\implies\alpha=\frac{a}{r}$. Let the radius of the truck's wheels be $r_t$ and the radius of the trailer's wheels be $r_{tr}$. Given that the trailer's wheels have a larger diameter (so larger radius, $r_{tr}>r_t$) and $a$ is the same for both (from part a, $a_t = a_{tr}$). Then $\alpha=\frac{a}{r}$, so $\alpha_t=\frac{a}{r_t}$ and $\alpha_{tr}=\frac{a}{r_{tr}}$. Since $r_{tr}>r_t$, $\alpha_{tr}<\alpha_t$. So the angular acceleration of the trailer's wheels is less than the angular acceleration of the truck's wheels.

### Question 5:
# Explanation:
## Step 1: Relationship between linear velocity, angular velocity, and radius
The relationship between linear velocity $v$, angular velocity $\omega$, and radius $r$ is $v = r\omega\implies\omega=\frac{v}{r}$. The number of revolutions $N$ is related to the angular displacement $	heta$ by $N=\frac{	heta}{2\pi}$, and $	heta=\omega t=\frac{v}{r}t$. Also, the distance traveled $s = vt$. And the number of revolutions can also be expressed as $N=\frac{s}{2\pi r}$ (since the circumference of the wheel is $2\pi r$, and the number of revolutions is the total distance divided by the circumference).
## Step 2: Effect of larger diameter (larger radius) on angular velocity
If we replace the tires with larger diameter (so larger $r$), and the speed $v$ is the same, from $\omega=\frac{v}{r}$, as $r$ increases, $\omega$ (angular velocity) decreases.
## Step 3: Effect of larger diameter on number of revolutions
The number of revolutions $N=\frac{s}{2\pi r}$. Since $s = vt$ (same speed and same distance, so $s$ is the same) and $r$ increases, $N$ (the number of revolutions) will decrease. Because as the radius of the wheel increases, the circumference ($2\pi r$) increases, so for the same distance traveled, the number of times the wheel needs to rotate (number of revolutions) decreases.

### Answers:
#### Question 1:
a. Second hand: $\boldsymbol{- 377}$ rad (or $\boldsymbol{3.77	imes 10^{2}}$ rad with three significant digits, considering direction)
b. Minute hand: $\boldsymbol{- 6.28}$ rad
c. Hour hand: $\boldsymbol{- 0.524}$ rad

#### Question 2:
a. Angular displacement in 3 min: $\boldsymbol{18.8}$ rad (or $6\pi$ rad)
b. Angular velocity: $\boldsymbol{6.28}$ rad/min (or $2\pi$ rad/min)
c. Angular acceleration is negative. Because the toy was rotating counter - clockwise (positive angular velocity) and when turned off, it will slow down, so the angular velocity is decreasing. The angular acceleration is in the opposite direction of the angular velocity (clockwise), so it is negative.

#### Question 3:
The diameter of the wheels is $\boldsymbol{\approx0.707}$ m (or 70.7 cm)

#### Question 4:
a. The linear acceleration of the trailer is equal to that of the truck.
b. The angular acceleration of the trailer's wheels is less than that of the truck's wheels.

#### Question 5:
- Angular velocity: Decreases (because $\omega=\frac{v}{r}$, $v$ constant, $r$ increases)
- Number of revolutions: Decreases (because $N=\frac{s}{2\pi r}$, $s$ constant, $r$ increases)

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QUESTION IMAGE

Question

Explanation:

Question 1:

Part a: Second Hand

Step 1: Determine the time and full rotation angle

Step 2: Calculate the number of full rotations

Step 3: Calculate the angular displacement

Step 4: Final calculation

Part b: Minute Hand

Step 1: Time and full rotation angle

Step 2: Angular displacement

Part c: Hour Hand

Step 1: Time and full rotation angle

Step 2: Angular displacement

Question 2:

Part a: Angular Displacement in 3 min

Answer:

Question 1:

Question 2:

Question 3:

Question 4:

Question 5: