First slide

Vertical circular motion

Question

A stone of mass 1kg is tied to one end of a string of length 0.5 m. It is whirled in a vertical circle. If the maximum tension in the string is 58.8N, the velocity at the top is

Moderate

A

1.82 ms^–1
B

2.2ms^–1
C

3.26 ms^–1
D

2.87 ms^–1

Solution

$\large T = mg + \frac{{m{v^2}}}{r} \Rightarrow 58.8 = 9.8 + \frac{{(1){v^2}}}{{0.5}} \Rightarrow 49 = \frac{{{v^2}}}{{0.5}}$
$\large {V^2} = \frac{{49}}{2} \Rightarrow V = \frac{7}{{\sqrt 2 }}\;m/s$
Conserving energy between top and the lowest positions.
1/2 mu² + mg.2l = 1/2 mV²
⇒ u² = V² - 4 gl
⇒ u² = 49 / 2 -4 X 9.8 X 0.5
⇒ u = 2.2 m/s

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