A thin circular ring of mass M and radius R is rotating about its axis with angular velocity omega. What is the torque required to stop it in t seconds?

A. MR^2 t — The torque is equal to the rate of change of angular momentum. For a ring, the moment of inertia is M R squared, so the change in angular momentum is M R squared omega, divided by time t.

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$A thin circular ring of mass M and radius R is rotating about its axis with angular velocity ω . What is the torque required to stop it in t seconds?$

A. $\frac{M R ^{2} ω}{t}$
B. $\frac{M R ^{2} ω}{2 t}$
C. $\frac{2 M R ^{2} ω}{t}$
D. $\frac{M R ^{2} ω}{4 t}$

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Correct answer: A — $\frac{M R ^{2} ω}{t}$

Solution

Torque τ = \frac{Δ L}{Δ t} . For a ring, I = M R^{2} . Thus, τ = \frac{I ω - 0}{t} = \frac{M R ^{2} ω}{t} .

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A thin circular ring of mass M and radius R is rotating about its axis with angular velocity ω. What is the torque required to stop it in t seconds?

Solution

Attempt this question & track your score

More Rotational Motion practice questions

$A thin circular ring of mass M and radius R is rotating about its axis with angular velocity ω . What is the torque required to stop it in t seconds?$