Question

A cycle is placed upside down while its owner repairs the tyre on the rear wheel. His friend spins the front wheel of radius

R

and observes that drops of water fly off tangentially in upward direction when the drops are at the same level as the centre of the wheel. He measures the height reached by drops moving vertically as shown in the figure. A drop that breaks loose from the tyre on one turn rises a distance

h_{1}

above the tangent point. A drop that breaks loose on the next turn rises a distance

h_{2} < h_{1}

above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. From this information, determine the magnitude of the average angular acceleration of the wheel.

Accepted Answer

The mechanical energy is conserved in the drop-earth system. At the instant it comes off the wheel, the first drop has a velocity $v_{1}$ directed upward. The magnitude of this velocity is found from conservation of energy,

$Δ K . E . + Δ U = 0 \Rightarrow (0 - \frac{1}{2} m v_{1}^{2}) + m g h_{1} = 0$

It means the value of linear speed,

$v_{1} = \sqrt{2 g h_{1}}$ .

The angular velocity of the wheel at the instant the first drop leaves is,

$ω_{1} = \frac{v_{1}}{R} = \sqrt{\frac{2 g h_{1}}{R^{2}}}$ .

Similarly, for the second drop,

$v_{2} = \sqrt{2 g h_{2}}$ and $ω_{2} = \frac{v_{2}}{R} = \sqrt{\frac{2 g h_{2}}{R^{2}}}$ .

Using the third equation of motion,

${(ω_{2})}^{2} = {(ω_{1})}^{2} + 2 α ∆ θ$ .

Then, the angular acceleration of the wheel is,

$a = \frac{ω_{2}^{2} - ω_{1}^{2}}{2 Δ θ} = \frac{(\frac{2 g h}{R^{2}}) - (\frac{2 g h_{1}}{R^{2}})}{2 (2 π)} = \frac{g (h_{2} - h_{1})}{2 π R^{2}}$ .

B M Sharma Solutions for Chapter: Circular Motion, Exercise 14: ILLUSTRATION 9.14

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 14: ILLUSTRATION 9.14

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 14: ILLUSTRATION 9.14 with Hints & Solutions

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