Question

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72 rev/s , friction in the bearings causes the wheel to stop in just 19 s .If the moment of inertia of the wheel about its axle is 0.30kg⋅m2, what is the magnitude of the frictional torque?

Answer 1

Answer:

0.0714 Nm

Explanation:

0.72 rev/s = 0.72 * 2π = 4.52 rad/s

Since it takes 19s to completely decelerate from 4.52 rad/s to 0 rad/s, we can calculate the angular deceleration caused by friction

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 - 4.52}{19} = -0.238 rad/s^2$

According to Newton's 2nd law, the magnitude of the frictional torque is the product of its acceleration and moment of inertia

$T = I\alpha = 0.3 * 0.238 = 0.0714 Nm$

Answer 2

The required magnitude of the frictional torque is 0.0714 N-m.

Given data:

The revolution rate is, n = 0.72 rev/s.

The time interval to stop is, t = 19 s.

The moment of inertia of wheel is, $I = 0.30 \;\rm kg.m^{2}$.

According to Newton's 2nd law, the magnitude of the frictional torque is the product of its acceleration and moment of inertia. Then,

$T = I \times \alpha$

Here, $\alpha$ is the angular acceleration. And its value is calculated as,

$\alpha = \dfrac{\omega}{t} \\\\\alpha = \dfrac{2 \pi n}{t} \\\\\alpha = \dfrac{2 \pi \times 0.72 }{19}\\\\\alpha = 0.238 \;\rm rad/s^{2}$

Then, the magnitude of the frictional torque is calculated as,

$T = 0.30 \times 0.238\\T = 0.0714 \;\rm N.m$

Thus, we can conclude that the required magnitude of the frictional torque is 0.0714 N-m.

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