A flywheel with a moment of inertia of 5.0 kg-m^2 is rotated at a speed of 60 rad/s . Due to friction at the axle, it comes to rest in 5.0 minutes . Determine the average torque of the friction.

Correct answer: C. 1.0 N-m Given moment of inertia, I = 5.0 kg-m^2 Initial angular velocity, ω_i = 60 rad/s Final angular velocity, ω_f = 0 rad/s Time taken, t = 5.0 minutes = 300 s Using the equation of rotational kinematics, ω_f = ω_i + α t 0 = 60 + α (300) α = -0.2 rad/s^2 The magnitude of the average torque of friction is given by = I |α| = 5.0 × 0.2 = 1.0 N-m

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Rotational Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rotational Mechanics

A flywheel with a moment of inertia of 5.0 kg-m^2 is rotated at a...

A flywheel with a moment of inertia of

5.0 \text{ kg-m}^2

is rotated at a speed of

60 \text{ rad/s}

. Due to friction at the axle, it comes to rest in

5.0 \text{ minutes}

. Determine the average torque of the friction.

Options

A $12.0 \text{ N-m}$
B $0.2 \text{ N-m}$
C $1.0 \text{ N-m}$
D $5.0 \text{ N-m}$

Correct answer

C. $1.0 \text{ N-m}$

Step-by-step solution

Given moment of inertia,

I = 5.0 \text{ kg-m}^2

Initial angular velocity,

\omega_i = 60 \text{ rad/s}

Final angular velocity,

\omega_f = 0 \text{ rad/s}

Time taken,

t = 5.0 \text{ minutes} = 300 \text{ s}

Using the equation of rotational kinematics,

\omega_f = \omega_i + \alpha t

0 = 60 + \alpha (300)

\alpha = -0.2 \text{ rad/s}^2

The magnitude of the average torque of friction is given by

\tau = I |\alpha|

\tau = 5.0 \times 0.2 = 1.0 \text{ N-m}

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