Question

A 0.16kg stone attached to a string of length I = 0.22m, is whirled in a horizontal circle with a constant velocity of 4.0m/s. a) Calculate the radial i.e the centripetal acceleration of the stone. b) The tension in the string while the stone is rotating. c) The horizontal force on the stone.

Answer 1

PART A)

`a_c = 72.7 m/s^2`

PART B)

T = 11.6 N

PART C)

`F_(horizontal) = 11.6 N`

Step-by-step explanation:

PART A)

Centripetal acceleration is given by

`a_c = (v^2)/(R)`

now we have

`a_c = (4^2)/(0.22)`

`a_c = 72.7 m/s^2`

PART B)

Here Tension force of string is providing centripetal force

so we can say

`T = ma_c`

`T = 0.16(72.7) = 11.64 N`

PART C)

Force on the stone in horizontal direction is only tension force

so here we have

`F_(horizontal) = T = 11.64 N`