Question

A uniform circular ring of mass m and radius R is placed freely on a horizontal smooth surface as shown in
figure. A particle of mass m is connected to the circumference of the ring with a massless string. The particle is imparted velocity v perpendicular to length of string as shown. If T is tension in the string just
after the particle imparted velocity ,then Acceleration of point P at this iñstant , is​

Answer 1

HOPE THIS HELPS YOU....

Answer 2

→ Hey Mate,

→ Given Question:-

→ A uniform circular ring of mass m and radius R is placed freely on a horizontal smooth surface as shown in

figure. A particle of mass m is connected to the circumference of the ring with a massless string. The particle is imparted velocity v perpendicular to length of string as shown. If T is tension in the string just

after the particle imparted velocity ,then Acceleration of point P at this iñstant , is?

→ Solution:-

→ T = m.a₀ (∵  Newton's first law )( Eq -1)

→ Net torque ( ∑Tc) = Ir

→ T.R = MR² .∝ ( ∵ Here R and R² gets cancelled)  

→ T = mR∝   ( Equation 2 )  

→ Then , solving equation 1 and 2

→  mR∝ = ma₀ ( ∵ here m and m gets cancel)

→ αо = R∝  is the solution.

→ ∵ ∝ = angular acceleration.

→ Now ,

→ αp bar ⇒ = α₀bar⇒ + αt bar⇒

→                 = R∝         +   R∝

→ αp bar ⇒ = 2R∝

→ ∵ R∝ = T / m

→ αp bar ⇒ = 2T / m is the solution.

hope it was helpful

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