Question

A ball of mass 50 g tied to the end of a 50 cm inextensible string is whirled around in a vertical circle. Find the tension in the string when the ball is at the top of the circle. Take
g=10m=s2

Answer 1

Answer on Question #40803, Physics, Mechanics

A ball of mass $50\mathrm{g}$ tied to the end of a $50~\mathrm{cm}$ inextensible string is whirled around in a vertical circle. Find the tension in the string when the ball is at the top of the circle. Take $\mathrm{g} = 10\mathrm{m}\cdot \mathrm{s}^2$.

Solution:

Given:

$$m = 50\mathrm{g} = 50 \cdot 10^{-3}\mathrm{kg},$$ $$r = 50\mathrm{cm} = 0.5\mathrm{m},$$ $$g = 10\mathrm{m}/\mathrm{s}^2,$$ $$T = ?$$

If you have a ball on the end of a string and you swing it in a vertical circle the "centripetal force" or the forces causing the acceleration will be a combination of the tension from the string and gravity.

The Tension and Weight are the forces causing the acceleration. The ball is also moving in a circle so at the highest and lowest points

Tension+Weight=CentripetalForce.

Hence,

$$F_{net} = ma = \frac{mv^2}{r}$$ $$T + W = ma$$

Thus,

$$T = ma - W = \frac{mv^2}{r} - mg = m\left(\frac{v^2}{r} - g\right)$$ $$T = m\left(\frac{v^2}{r} - g\right)$$

The tension depends on speed of ball $v$.

You need to know the speed of ball.

If $v = \sqrt{gr}$, then $T = 0$.

If $v > \sqrt{gr}$, then $T > 0$.

**Answer.** $T = m\left(\frac{v^2}{r} - g\right)$.