Question

$FIGUREQ8.3$is a bird’s-eye view of particles on strings moving in horizontal circles on a tabletop. All are moving at the same speed. Rank in order, from largest to smallest, the tensions $T_{a}to{T}_d$. Give your answer in the form$a>b=c>d$and explain your ranking.

Short answer

The tensions are ranked in the order:$T_c>T_b>T_d=T_a$

Step-by-step solution

Given information

The tensions at mass $m$with velocity $v$and radius$r$of circular motion $T_{a}to{T}_d$

Centripetal force equal to tension acting on the string.

The free-body diagram shows that the downward gravitational force is balanced by an upward component of the tension, leaving the radial component of the tension to cause the centripetal acceleration. Newton’s second law is

$\sum _{}{F}_r=T=\frac{m{v}^2}{r}$

The above tension equation is applied for the given cases of free body diagram of FIGURE Q8.3

Explanation

case (a): The tension at mass m with velocity v and radius of circular motion gives

$T_a=\frac{m{v}^2}{r}$

case (b): The tension at mass m with velocity v and radius of circular motion gives

$T_b=\frac{m{v}^2}{2r}=0.5\frac{m{v}^2}{r}$

case (c): The tension at mass m with velocity v and radius of circular motion gives

$T_c=\frac{2m{v}^2}{r}$

case (d): The tension at mass m with velocity v and radius of circular motion gives

$T_d=\frac{2m{v}^2}{2r}=\frac{m{v}^2}{r}$

The Tensions of each string for different cases are ranked in the order

$T_c>T_b>T_d=T_a$