Chapter 1: Q15E (page 330)

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12 rev/s after 2.00 s. What is the moment of inertia of the wheel?

Expert verified

The moment of inertia of the wheel is, $I = 0.255 k g \cdot m^{2}$ .

Step by step solution

We have the given data:

The radius of the wheel $(r)$ =0.120 m.

Force applied to the rim of the wheel $(F)$ =80.0 N.

Angular speed of the wheel $(ω_{z}) =$ 12.0 rev/s =75.40 rad/s.

The system is initially at rest.

Time interval during which the revolutions take place is,

$Δ t = 2.00$ s.

If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration $α$ , where,

$\sum τ_{e x t} = I α \dots \dots (1)$

First, we will calculate angular acceleration which is given by,

$ω_{z} = ω_{0 z} + α_{z} t$ .

Since the system starts from the rest, we have, $ω_{0 z} = 0$ .

Therefore,

ω_{z} = α_{z} t \Rightarrow α_{z} = \frac{ω_{z}}{t} \Rightarrow α_{z} = \frac{75.40}{2.00} \Rightarrow α_{z} = 37.70 r a d / s^{2}

From $(1)$ , the net torque is given by,

\sum τ_{z} = I α_{z} \Rightarrow I = \frac{\sum τ_{z}}{α_{z}} \Rightarrow I = \frac{F r}{α_{z}} \Rightarrow I = \frac{(80.0) (0.120)}{(37.70)} \Rightarrow I = 0.255 k g \cdot m^{2}

Hence, the moment of inertia of the wheel is, $I = 0.255 k g \cdot m^{2}$ .

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