Question

A ball hung at the bottom end of a string with a length of 1.2 m undergoes a horizontal circular motion with a radius of 0.8 m. (a) What force acts as the centripetal force for the circular motion of the ball? (b) What is the magnitude of centripetal acceleration? (c) What is its angular velocity? (d) How long (in minutes) does the ball take to finish 100 revolution? Note: the friction can be neglected in this question.

Answer 1

Answer on Question #75699-Physics-Other

A ball hung at the bottom end of a string with a length of $1.2\,\mathrm{m}$ undergoes a horizontal circular motion with a radius of $0.8\,\mathrm{m}$.

(a) What force acts as the centripetal force for the circular motion of the ball?

(b) What is the magnitude of centripetal acceleration?

(c) What is its angular velocity?

(d) How long (in minutes) does the ball take to finish 100 revolution? Note: the friction can be neglected in this question.

Solution

(a) The horizontal projection of tension force.

(b)

$$a = \frac{v^2}{r} = g \tan \theta$$ $$\theta = \sin^{-1} \frac{r}{l} = \sin^{-1} \frac{0.8}{1.2} = 41.8{}^\circ$$ $$a = 9.8 \tan 41.8{}^\circ = 8.8 \frac{m}{s^2}$$

(c)

$$\omega = \sqrt{\frac{a}{r}} = \sqrt{\frac{9.8 \tan 41.8{}^\circ}{0.8}} = 3.3 \frac{rad}{s}$$

(d)

$$t = 100T = 100 \frac{2\pi}{\omega} = 100 \frac{2\pi}{3.3} = 190\,s = 3.2\,\mathrm{min}$$

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