Question

A mass m hangs with the help of string wrapped around a pulley on a frictionless bearing . the pulley has mass m and radius R . Assuming pulley to be a perfect uniform circular disc , the acceleration of the mass m , if the string does not slip on the pulley ,is [AIEEE 2011] ​

Answer 1

Answer:

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2g/3

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$mg−T=ma\\</p><p>TR= (m{R}^{2}a)/2\\</p><p>T= mRa/2 = ma/2\\</p><p>3ma/2=mg\\</p><p>a= 2g/3$

Answer 2

Explanation :- For the motion of the block

• mg -T = ma ______(1)
• For the rotation of the pulley .
• ζ= TR = Iα
• we know centre of mass of circular disc about the perpendicular axis is 1/2mr²
• => TR= 1/2mR²α
• As string doesn't slip on the pulley
• a = Rα
• ∴T = 1/2ma______(2)

From (1) and (2)

• mg -T = ma
• mg-1/2ma =ma
• mg = 3ma/2
• a = 2g/3Answer

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