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

Answer:

PART A)

$a_c = 72.7 m/s^2$

PART B)

T = 11.6 N

PART C)

$F_{horizontal} = 11.6 N$

Explanation:

PART A)

Centripetal acceleration is given by

$a_c = \frac{v^2}{R}$

now we have

$a_c = \frac{4^2}{0.22}$

$a_c = 72.7 m/s^2$

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

$F_{horizontal} = T = 11.64 N$