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,

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

$F_{cent} = -F_g +F_T$

$F_T = F_{cent}- F_g$

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

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

$F_T = 113.8899N$

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