Question

A 0.75 ball is tied to a 0.50 m long string and whirled in a vertical circle at a constant speed of 5m/s. What is the tension in the string at the top and the bottom of the circle?

Answer 1

Answer on Question #84882, Physics / Mechanics | Relativity

A 0.75 (kg) ball is tied to a 0.50 m long string and whirled in a vertical circle at a constant speed of 5ms. What is the tension in the string at the top and the bottom of the circle

Solution:

We draw a free body force diagram of the mass. At the top of the circle the two forces acting on a ball are its weight $W = mg$ and tension from the string $T$. Both of them act vertically

downwards. The centripetal force $F$ acting towards the center of the circle is

$$F = \frac{m v^2}{r} = T_{top} + mg$$

where $m$ is mass, $v$ is velocity of the ball and $r$ is radius, $g = 10\,m/s^2$ is the gravitational acceleration. Then for tension in the string at the top we get

$$T_{top} = \frac{m v^2}{r} - mg = m \left(\frac{v^2}{r} - g\right) = 0.75 \left(\frac{s^2}{0.5} - 10\right) = 0.75(50 - 10) = 30\,N$$

At the bottom of the circle the weight acts vertically downwards, whereas the tension from the string acts vertically upwards so that the centripetal force is

$$F = \frac{m v^2}{r} = T_{bot} - mg$$

Then for $T_{bot}$ we get

$$T_{bot} = \frac{m v^2}{r} + mg = m \left(\frac{v^2}{r} + g\right) = 0.75 \left(\frac{s^2}{0.5} + 10\right) = 0.75(50 + 10) = 45\,N$$

Answer: the tension in the string at the top and the bottom of the circle are

$$T_{top} = 30\,N, \quad T_{bot} = 45\,N$$

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