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A wheel has moment of inertia 5 xx 10^(-...

A wheel has moment of inertia `5 xx 10^(-3) kg m^(2)` and is making `20 "rev" s^(-1)`. The torque needed to stop it in `10 s` is….. `xx 10^(-2) N-m`

A

`2pi`

B

`2.5pi`

C

`4pi`

D

`4.5pi`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Convert the angular velocity from revolutions per second to radians per second. The initial angular velocity \( \omega_0 \) is given as 20 revolutions per second. We need to convert this to radians per second using the conversion factor \( 2\pi \) radians per revolution. \[ \omega_0 = 20 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 40\pi \, \text{rad/s} \] ### Step 2: Calculate the angular retardation needed to stop the wheel in 10 seconds. The wheel needs to stop in \( t = 10 \, \text{s} \). We can calculate the angular retardation \( \alpha \) using the formula: \[ \alpha = \frac{\Delta \omega}{t} \] Since the final angular velocity \( \omega_f = 0 \, \text{rad/s} \), we have: \[ \Delta \omega = \omega_f - \omega_0 = 0 - 40\pi = -40\pi \, \text{rad/s} \] Now substituting into the equation: \[ \alpha = \frac{-40\pi}{10} = -4\pi \, \text{rad/s}^2 \] ### Step 3: Calculate the torque required to produce this angular retardation. The torque \( \tau \) is related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by the equation: \[ \tau = I \cdot \alpha \] Given that the moment of inertia \( I = 5 \times 10^{-3} \, \text{kg m}^2 \), we can substitute the values: \[ \tau = 5 \times 10^{-3} \cdot (-4\pi) \] Calculating this gives: \[ \tau = -20\pi \times 10^{-3} \, \text{N m} \] Since we are interested in the magnitude of torque needed to stop the wheel, we take the absolute value: \[ \tau = 20\pi \times 10^{-3} \, \text{N m} \] ### Step 4: Express the answer in the required format. The torque needed to stop the wheel in \( 10 \, \text{s} \) is: \[ \tau = 2\pi \times 10^{-2} \, \text{N m} \] ### Final Answer: The torque needed to stop the wheel in \( 10 \, \text{s} \) is \( 2\pi \times 10^{-2} \, \text{N m} \). ---
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