# a mass m moves in a circle on a smooth horizontal plane with velocity v0

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## A mass, m , moves in a circle on a smooth horizontal plane with velocity, V0 , at a radius, R0 . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R02 . The final value of the kinetic energy is

Click here👆to get an answer to your question ✍️ A mass, m , moves in a circle on a smooth horizontal plane with velocity, V0 , at a radius, R0 . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R02 . The final value of the kinetic energy is

A mass, m, moves in a circle on a smooth horizontal plane with velocity, VQuestion 0 , at a radius, R 0

. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius

2 R 0

. The final value of the kinetic energy is

**A**2mv

0 2

**B**

2 1 mv 0 2

**C**mv

0 2

**D**

4 1 mv 0 2 Medium NEET Open in App Solution Verified by Toppr

Correct option is A)

The angular momentum of mass about center will remain constant as the torque of tension is zero.mv ∘ r ∘ =mv ′ 2 r ∘ v ′ =2v ∘

Final value of K.E.=

2 1 m(2v ∘ ) 2 =2mv ∘ 2 Video Explanation

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Question A mass m

moves in a circle on a smooth horizontal plane with velocity

v 0 at a radius R 0

. The mass is attached to a string which passes through a smooth hole in the plane as shown.

The tension in the string is increased gradually and finally

m moves in a circle of radius R 0 / 2

. Then, the final value of the kinetic energy of particle is

Open in App Solution

The correct option is **D**

2 m v 0 2

Since no external torque is acting on the system, thus according to law of conservation of angular momentum,

Initial angular momentum = Final angular momentum

m v 0 R 0 = m v r ⇒ m v 0 R 0 = m v R 0 2 ( ∵ New radius , r = R 0 2 ) ⇒ v = 2 v 0 New kinetic energy = 1 2 m v 2 = 2 m v 2 0

COM in Collision and Explosion

Standard XII Physics

Suggest Corrections 10

## A mass m moves in a circles on a smooth horizontal plane with velocity v(0) at a radius R(0). The mass is atteched to string which passes through a smooth hole in the plane as shown. The tension in string is increased gradually and finally m moves in a cricle of radius (R(0))/(2). the final value of the kinetic energy is

( c) Initial angular momentum L(initial) = mv0 R Initial angular momentum L(initial) = mv ( R)/(2) Conservation of angular momentum m v0 R0 =m mv (R0)/(2) rArr v = 2 v0 KE = (1)/(2) mv^2 = 2 mv0^2.

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A mass m moves in a circles on...

A mass m m

moves in a circles on a smooth horizontal plane with velocity

v 0 v0 at a radius R 0 R0

. The mass is atteched to string which passes through a smooth hole in the plane as shown.

The tension in string is increased gradually and finally

m m

moves in a cricle of radius

R 0 2 R02

. the final value of the kinetic energy is

Updated On: 27-06-2022

00 : 30

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Text Solution Open Answer in App A m v 2 0 mv02 B 2m v 2 0 2mv02 C 1 4 m v 2 0 14mv02 D 1 2 m v 2 0 12mv02 Answer

The correct Answer is B

Solution

( c) Initial angular momentum

L ∈itial =m v 0 R L∈itial=mv0R

Initial angular momentum

L ∈itial =mv R 2 L∈itial=mvR2

Conservation of angular momentum

m v 0 R 0 =mmv R 0 2 ⇒v=2 v 0 mv0R0=mmvR02⇒v=2v0 KE= 1 2 m v 2 =2m v 2 0 KE=12mv2=2mv02 .

Answer

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Guys, does anyone know the answer?