Question

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

Answer 1

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.