Linear and Angular Velocity Worksheet for IB Math I

Problem 1:

A large pulley 20 cm in diameter drives a smaller pulley 6 cm in diameter by a belt that goes over the rim of each. The large pulley has an angular velocity of 150 radians per minute.

#### Part (a): What is the linear velocity of the large pulley’s rim?

The formula for linear velocity \( v \) is given by:
\[
v = r \cdot \omega
\]
where:
- \( r \) is the radius of the pulley,
- \( \omega \) is the angular velocity.

For the large pulley:
- Diameter = 20 cm, so radius \( r = \frac{20}{2} = 10 \) cm,
- Angular velocity \( \omega = 150 \) radians per minute.

Substitute these values into the formula:
\[
v = 10 \cdot 150 = 1500 \text{ cm/min}
\]

Thus, the linear velocity of the large pulley’s rim is:
\[
\boxed{1500 \text{ cm/min}}
\]

#### Part (b): What is the linear velocity of the smaller pulley’s rim?

Since the belt drives both pulleys and does not slip, the linear velocity of the belt (and thus the rims of both pulleys) must be the same. Therefore, the linear velocity of the smaller pulley’s rim is also:
\[
\boxed{1500 \text{ cm/min}}
\]

#### Part (c): What is the angular velocity of the smaller pulley?

The formula relating linear velocity \( v \), radius \( r \), and angular velocity \( \omega \) is:
\[
v = r \cdot \omega
\]
Rearranging for \( \omega \):
\[
\omega = \frac{v}{r}
\]

For the smaller pulley:
- Linear velocity \( v = 1500 \) cm/min (from part (b)),
- Radius \( r = \frac{6}{2} = 3 \) cm.

Substitute these values into the formula:
\[
\omega = \frac{1500}{3} = 500 \text{ radians/min}
\]

Thus, the angular velocity of the smaller pulley is:
\[
\boxed{500 \text{ radians/min}}
\]

---

Problem 2:

A large gear of diameter 30 cm is revolving at 45 rpm (revolutions per minute). It drives a smaller gear of diameter 8 cm.

#### Part (a): How many radians per minute is the large gear turning?

The relationship between revolutions per minute (rpm) and radians per minute is:
\[
\text{Angular velocity in radians per minute} = \text{rpm} \cdot 2\pi
\]

For the large gear:
- rpm = 45.

Substitute this value:
\[
\omega = 45 \cdot 2\pi = 90\pi \text{ radians/min}
\]

Thus, the large gear is turning at:
\[
\boxed{90\pi \text{ radians/min}}
\]

#### Part (b): What is the linear velocity of the teeth on the large gear?

The formula for linear velocity \( v \) is:
\[
v = r \cdot \omega
\]

For the large gear:
- Diameter = 30 cm, so radius \( r = \frac{30}{2} = 15 \) cm,
- Angular velocity \( \omega = 90\pi \) radians per minute.

Substitute these values:
\[
v = 15 \cdot 90\pi = 1350\pi \text{ cm/min}
\]

Thus, the linear velocity of the teeth on the large gear is:
\[
\boxed{1350\pi \text{ cm/min}}
\]

#### Part (c): What is the linear velocity of the teeth on the smaller gear?

Since the gears are connected by a belt that does not slip, the linear velocity of the teeth on both gears must be the same. Therefore, the linear velocity of the teeth on the smaller gear is also:
\[
\boxed{1350\pi \text{ cm/min}}
\]

#### Part (d): How many radians per minute is the smaller gear turning?

The formula relating linear velocity \( v \), radius \( r \), and angular velocity \( \omega \) is:
\[
v = r \cdot \omega
\]
Rearranging for \( \omega \):
\[
\omega = \frac{v}{r}
\]

For the smaller gear:
- Linear velocity \( v = 1350\pi \) cm/min (from part (c)),
- Radius \( r = \frac{8}{2} = 4 \) cm.

Substitute these values:
\[
\omega = \frac{1350\pi}{4} = 337.5\pi \text{ radians/min}
\]

Thus, the smaller gear is turning at:
\[
\boxed{337.5\pi \text{ radians/min}}
\]

#### Part (e): How many revolutions per minute is the smaller gear turning?

The relationship between angular velocity in radians per minute and revolutions per minute is:
\[
\text{rpm} = \frac{\omega}{2\pi}
\]

For the smaller gear:
- Angular velocity \( \omega = 337.5\pi \) radians per minute.

Substitute this value:
\[
\text{rpm} = \frac{337.5\pi}{2\pi} = 168.75
\]

Thus, the smaller gear is turning at:
\[
\boxed{168.75 \text{ rpm}}
\]

---

Problem 3:

An LP (long playing) record rotates at 33 1/3 revolutions per minute (rpm).

#### Part (a): How many radians per minute is this?

The relationship between revolutions per minute (rpm) and radians per minute is:
\[
\text{Angular velocity in radians per minute} = \text{rpm} \cdot 2\pi
\]

For the LP record:
- rpm = 33 1/3 = \( \frac{100}{3} \).

Substitute this value:
\[
\omega = \frac{100}{3} \cdot 2\pi = \frac{200\pi}{3} \text{ radians/min}
\]

Thus, the angular velocity is:
\[
\boxed{\frac{200\pi}{3} \text{ radians/min}}
\]

#### Part (b): Find the angular velocity at the needle when it is just starting the record, about 14.5 cm from the center.

The angular velocity of a point on a rotating object is the same regardless of its distance from the center. Therefore, the angular velocity at the needle (14.5 cm from the center) is the same as the angular velocity of the entire record:
\[
\boxed{\frac{200\pi}{3} \text{ radians/min}}
\]

---

Final Answers:

1. (a) \( \boxed{1500 \text{ cm/min}} \)
(b) \( \boxed{1500 \text{ cm/min}} \)
(c) \( \boxed{500 \text{ radians/min}} \)

2. (a) \( \boxed{90\pi \text{ radians/min}} \)
(b) \( \boxed{1350\pi \text{ cm/min}} \)
(c) \( \boxed{1350\pi \text{ cm/min}} \)
(d) \( \boxed{337.5\pi \text{ radians/min}} \)
(e) \( \boxed{168.75 \text{ rpm}} \)

3. (a) \( \boxed{\frac{200\pi}{3} \text{ radians/min}} \)
(b) \( \boxed{\frac{200\pi}{3} \text{ radians/min}} \)