Question

A wheel of 40cm radius rotates on a stationary rim. It is uniformly speeded up from a rest to a speed of 900 rpm. In a time of 20 sec.
A. Find the constant acceleration.
B. The tangential accelerating of the rim.

2. A pulley of 5cm radius of a motor is turning at 30 revolution per second. And slow down uniformly to 20 revolution per second in 2 second. Calculate
A. The angular acceleration of the motor
B. The number of revolution it made this time
C. The length of belt it winds in this time

Answer 1

Question #77660, Physics / Other

A wheel of 40cm radius rotates on a stationary rim. It is uniformly speeded up from a rest to a speed of 900 rpm. In a time of 20 sec.

A. Find the constant acceleration.

B. The tangential accelerating of the rim.

Solution

a.

$$a _ {N} = r \omega^ {2} = 0.4 \left(900 \frac {2 \pi}{60}\right) ^ {2} = 3553 \frac {m}{s ^ {2}}$$

b.

$$a _ {\tau} = \frac {r \omega}{t} = \frac {0.4 \left(900 \frac {2 \pi}{60}\right)}{20} = 1.9 \frac {m}{s ^ {2}}$$

2. A pulley of 5cm radius of a motor is turning at 30 revolution per second. And slow down uniformly to 20 revolution per second in 2 second. Calculate

A. The angular acceleration of the motor

B. The number of revolution it made this time

C. The length of belt it winds in this time

Solution

a.

$$\alpha = \frac {\Delta \omega}{t} = \frac {\left(20 \frac {2 \pi}{1}\right) - \left(30 \frac {2 \pi}{1}\right)}{2} = - 10 \pi \frac {rad}{s ^ {2}} = - 31 \frac {rad}{s ^ {2}}$$

b.

$$2 \alpha \theta = \omega _ {2} ^ {2} - \omega _ {1} ^ {2}$$ $$2 \alpha (2 \pi) n = \omega _ {2} ^ {2} - \omega _ {1} ^ {2}$$ $$n = \frac {1}{4 \pi (- 10 \pi)} \left(\left(20 \frac {2 \pi}{1}\right) ^ {2} - \left(30 \frac {2 \pi}{1}\right) ^ {2}\right) = 50 \text{ rev.}$$

c.

$$l = 2 \pi r n = 2 \pi (0.05) (50) = 16 \text{ m}$$

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