Question

A 2.2 kg object is whirled in a vertical circle whose radius is 1.0 m. If the time of one revolution is 0.97 s,
what is the tension in the string (assume uniform speed)
a) at the top?
b) at the bottom?

Answer 1

Answer:

The tension in the string at the top  =  $71 N$

The tension in the string at the top  =  $1.1 \times 10^{-2} N$

Explanation:

a) at the top  

$F_{cent} = F_g +F_T$ ----------------------(1)

Where,

$F_{cent}$ is the centripetal force

$F_g$ is the gravitational force

$F_T$  force due to tension

From (1)

$F_T = F_{cent} - F_g$

$F_T = \frac{ 4\pi^2 Rm}{T^2} -mg$

where

R is the radius

m is the mass

T is the time taken for one revolution

g is the acceleration due to gravity

On Substituting the values

$F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} -(2.2 \times9.8)$

$F_T$= 70.7259N

$F_T$ =71N

b) at the bottom

On Substituting the values

$F_{cent} = -F_g +F_T$

$F_T = F_{cent}- F_g$

$F_T = \frac{ \pi^2 Rm}{T^2} +mg$

$F_T = \frac{4 (3.14)^2 (1.0)}{(0.97)^2} +(2.2 \times9.8)$

$F_T = 113.8899N$

$F_T = 1.1 \times 10^{-2}$