Question

A net torque of 36 N.m acts on a wheel rotating about a fixed axis for 6 s. During this time the angular speed of the wheel increases from 0 to 12 rad/s. The applied force is then removed, and the wheel comes to rest in 75 s.
a. What is the moment of inertia of the wheel?
b. What is the magnitude of the frictional torque?
c. How many revolutions does the wheel make?

Answer 1

Answer on Question #42319-Physics-Mechanics-Kinematics

A net torque of 36 N.m acts on a wheel rotating about a fixed axis for 6 s. During this time the angular speed of the wheel increases from 0 to 12 rad/s. The applied force is then removed, and the wheel comes to rest in 75 s.

a. What is the moment of inertia of the wheel?

b. What is the magnitude of the frictional torque?

c. How many revolutions does the wheel make?

Solution

a. Angular acceleration

$$\alpha = \frac {\omega_ {2} - \omega_ {1}}{t _ {1}} = \frac {(1 2 - 0) \frac {\mathrm {r a d}}{\mathrm {s}}}{6 \mathrm {s}} = 2 \frac {\mathrm {r a d}}{\mathrm {s} ^ {2}}.$$ $$\mathrm {t o r q u e} = I \cdot \alpha \rightarrow I = \frac {\mathrm {t o r q u e}}{\alpha} = \frac {3 6 \mathrm {N} \cdot \mathrm {m}}{2 \frac {\mathrm {r a d}}{\mathrm {s} ^ {2}}} = 1 8 \mathrm {k g} \cdot \mathrm {m} ^ {2}.$$

b. The wheel comes to rest only due to the frictional torque $\tau_{friction}$

$$\tau_ {f r i c t i o n} = I \alpha^ {\prime}.$$

Angular deceleration

$$\alpha^ {\prime} = \frac {\omega_ {2} - \omega_ {3}}{t _ {2}} = \frac {(1 2 - 0) \frac {\mathrm {r a d}}{\mathrm {s}}}{7 5 \mathrm {s}} = 0. 1 6 \frac {\mathrm {r a d}}{\mathrm {s} ^ {2}}.$$ $$\tau_ {f r i c t i o n} = 1 8 k g \cdot m ^ {2} \cdot 0. 1 6 \frac {r a d}{s ^ {2}} = 2. 8 8 N \cdot m.$$

c. In time 6s

$$\theta_ {1} = \frac {1}{2} \alpha t _ {1} ^ {2} = 0. 5 \cdot 2 \cdot 6 ^ {2} = 3 6 r a d i a n s.$$

From average angular velocity formula

$$\theta_ {2} = \widetilde {\omega} t _ {2} = \frac {(1 2 - 0) \frac {\mathrm {r a d}}{\mathrm {s}}}{2} \cdot 7 5 \mathrm {s} = 4 5 0 \mathrm {r a d i a n s}.$$

Total angle traversed

$$\theta = \theta_ {1} + \theta_ {2} = 3 6 + 4 5 0 = 4 8 6 \mathrm {r a d i a n s}.$$

Number of revolutions

$$N = \frac {4 8 6}{2 \pi} \text{ revolutions} = 7 7. 3 \text{ revolutions}.$$

Answer: a. $18 \, \text{kg} \cdot \text{m}^2$; b. 2. $88 \, \text{N} \cdot \text{m}$; c. 77.3 revolutions.

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