A body of mass m is rotated uniform speed along a vertical circle with the of light string- If T-1- and T-2- are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively- then which of the following expressions is correct-T-2-T-1-6mgT-2-T-1-4mgT-2-T-1-2mgT-2-T-1-mg

Question

A body of mass m is rotated uniform speed along a vertical circle with the of light string- If T-1- and  T-2- are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively- then which of the following expressions is correct-T-2-T-1-6mgT-2-T-1-4mgT-2-T-1-2mgT-2-T-1-mg

Toppr Admin · Accepted Answer

Centripetal Force is- mv2rAt the highest point- T1-mv2r-x2212-mgAt the lowest point- T2-mv2r-mg-x2234-T2-x2212-T1-mv2r-mg-x2212-mv2r-x2212-mg-2mg

	Lower Point	Highest Point
a.	mg - $T_{1}$	mg + $T_{2}$
b.	mg + $T_{1}$	mg - $T_{2}$
c.	mg + $T_{1} - (m v_{1}^{2}) / R$	mg - $T + (m v_{1}^{2}) / R$
d.	mg - $T_{1} - (m v_{1}^{2}) / R$	mg + $T_{2} + (m v_{1}^{2}) / R$

Centripetal Force is: mv2rAt the highest point: T1=mv2r−mgAt the lowest point: T2=mv2r+mg∴T2−T1=mv2r+mg−(mv2r−mg)=2mg

Centripetal Force is: $\frac{m v^{2}}{r}$
At the highest point: $T_{1} = \frac{m v^{2}}{r} - m g$
At the lowest point: $T_{2} = \frac{m v^{2}}{r} + m g$
$∴ T_{2} - T_{1} = \frac{m v^{2}}{r} + m g - (\frac{m v^{2}}{r} - m g) = 2 m g$