Chapter 1: Q23E (page 330)

A $392 - N$ wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at $25.0 rad / s$ . The radius of the wheel is $0.600 m$ , and its moment of inertia about its rotation axis is $0.800 {MR}^{2}$ . Friction does work on the wheel as it rolls up the hill to a stop, a height h above the bottom of the hill; this work has absolute value $2600 J$ . Calculate h.

Short Answer

Expert verified

The height is 14 m.

Step by step solution

To mention the given data

We have

Weight of the wheel $(w) = 392 N$ .

Radius of the wheel $(R) = 0.600 m$ .

Angular speed $(ω_{0}) = 25.0 r a d / s$ .

Work done by the wheel $W_{f} =$ 2600 J

To find the height

Here we use energy conservation law:

The wheel has initially both rotational and translational energies, then it stopped at height h.

Therefore, we have,

$K_{t r} + K_{r o t} - W_{f} = P \Rightarrow \frac{1}{2} m v_{0}^{2} + \frac{1}{2} I ω_{0}^{2} - W_{f} = m g h$

Now, putting $v_{0} = ω_{0}, I = 0.8 m R^{2}$ and substituting values in above equation, we get,

localid="1667970909281" $\frac{1}{2} \frac{w}{g} R^{2} ω_{0}^{2} + (0.8) \frac{1}{2} \frac{w}{g} R^{2} ω_{0}^{2} - W_{f} = \frac{w}{g} g h \dots \dots (∵ m = \frac{w}{g}) \Rightarrow \frac{1}{2} (\frac{392}{9.80}) {(0.6)}^{2} {(25)}^{2} + (0.8) \frac{1}{2} (\frac{392}{9.80}) {(0.6)}^{2} {(25)}^{2} - 2600 = (392) h \Rightarrow h = \frac{\frac{1}{2} (\frac{392}{9.80}) {(0.6)}^{2} {(25)}^{2} + (0.8) \frac{1}{2} (\frac{392}{9.80}) {(0.6)}^{2} {(25)}^{2} - 2600}{392} \Rightarrow h = 14 m$

Hence, the height is 14 m.

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