A body of moment of inertia 2 kg m2 is rotated at a speed of 25 rad s–1. A tangential force at rim stops the wheel in10 s. Average torque of force
Question
A body of moment of inertia 2 kg m2 is rotated at a speed of 25 rad s–1. A tangential force at rim stops the wheel in10 s. Average torque of force
Solution
The problem involves the concept of rotational motion, specifically torque and angular momentum. Here are the steps to solve it:
Step 1: Identify the given values. Moment of inertia (I) = 2 kg m^2 Initial angular speed (ωi) = 25 rad/s Final angular speed (ωf) = 0 rad/s (since the wheel is stopped) Time (t) = 10 s
Step 2: Calculate the angular deceleration (α). We can use the formula for angular deceleration, which is α = (ωf - ωi) / t. Substituting the given values, we get α = (0 - 25) / 10 = -2.5 rad/s^2. The negative sign indicates that it's a deceleration.
Step 3: Calculate the average torque (τ). Torque can be calculated using the formula τ = I * α. Substituting the given values, we get τ = 2 kg m^2 * -2.5 rad/s^2 = -5 Nm.
The negative sign indicates that the torque is acting in the opposite direction to the initial rotation, which makes sense as it's stopping the wheel. So, the average torque of the force is 5 Nm.
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