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.

a) Determine the tension in the string when the ball is at the top of the circle;

b) at the bottom.

c) Consider the ball at some point other than the top or bottom. What can you say about the tension in the string at this point

Answer 1

Step-by-step 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=(mv^2)/(r)-mg`

`F=m((v^2)/(r)-g)`

`F=1\ lb* (((10\ ft/s)^2)/(2)-1\ lb* 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=(mv^2)/(r)+mg`

`F=m((v^2)/(r)+g)`

`F=1\ lb* (((10\ ft/s)^2)/(2)+1\ lb* 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((r-h)/(r))+(mv^2)/(r)`

`T=(m)/(r)(g(r-h)+v^2)`

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

`T=(m)/(r)(u^2-3gh+gr)`

Hence, this is the required solution.