Question

An object with a mass of 5 kg is swung in a vertical circle by a rope with a length of 0.67 m. The tension at the bottom of the circle is 88 Newtons. What is the tension, in Newtons, at the side of the circle, halfway between the top and bottom if the speed of the mass is the same at the bottom and side

Answer 1

Answer:

$T_2=39.5N$

Explanation:

From the question we are told that:

Mass $m=5kg$

Length $L=0.67m$

Tension $T=88N$

Generally the equation for Tension is mathematically given by

 $T = m * ( g + v^2 /l)$

Therefore

 $T_1 = m * ( g + \frac{v^2}{l})$

 $88 = 5 * ( 9.8 + \frac{v^2}{0.67})$

 $v^2=5.2$

 $v=2.4m/s$

The uniform velocity is

 $v=2.4m/s$

Therefore

The tension at the side of the circle halfway between the top and bottom is

 $T_2=5*\frac{2.3^2}{0.67}$

 $T_2=39.5N$