First slide

Torque

Question

The wheel on a grinder is a uniform 2 kg disc of 25 cm radius. It comes uniformly to rest from 1440 rpm in a time of 31.4 s. How large a friction torque slows its motion?

Moderate

A

0.1 Nm
B

0.3 Nm
C

0.2 Nm
D

0.4 Nm

Solution

$\begin{array}{l} ω = 1440 \times \frac{2 π}{60} = 48 π r a d / s^{2} \\ α = \frac{ω_{f} - ω_{i}}{t} = \frac{- 48 π}{31.4} = - 4.8 r a d / s^{2} \\ \begin{array}{l} I = \frac{1}{2} m r^{2} = \frac{1}{2} (2) {(0.25)}^{2} = 62.5 \times 10^{- 3} k g . m^{2} \\ τ = I α = (62.5 \times 10^{- 3}) (- 4.8) = 0.3 N m \end{array} \end{array}$

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