A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity √3 gr at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

Question

A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity √3 gr at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is (A) 4: 3 (B) 5: 4 (C) 1: 4 (D) 3: 2

Tardigrade Admin · Accepted Answer

Tension at highest point TH=(m v2/r)-m g=3 m g-m g=2 m g and tension at lowest point TL=(m v2/r)+m g Here, vL2=3 g r+2 g ⋅ 2 t=7 gr So, TL=7 m g+m g=8 mg Hence,(TH/TL)=(2 m g/8 mg)=(1/4)

Q. A particle tied to a string describes a vertical circular motion of radius $r$ continually. If it has a velocity $3 g r$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

4 : 3

5 : 4

1 : 4

3 : 2

1 : 2

Tension at highest point
$T_{H} = \frac{m v ^{2}}{r} - m g = 3 m g - m g = 2 m g$
and tension at lowest point
$T_{L} = \frac{m v ^{2}}{r} + m g$
Here, $v_{L}^{2} = 3 g r + 2 g \cdot 2 t = 7 g r$
So, $T_{L} = 7 m g + m g = 8 m g$
Hence, $\frac{T _{H}}{T _{L}} = \frac{2 m g}{8 m g} = \frac{1}{4}$

Q. A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity 3gr​ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is