Question

The pulley (uniform disc) shown in figure has, radius $10 cm$ and moment of inertia about its axis $I = 0.5 kg m^{2}$ ( $B$ and $A$ both move)

(a) Assuming all the plane surfaces are smooth and there is no slipping between pulley and string, calculate the acceleration of the mass $4 kg$ .

(b) The friction coefficient between the block $A$ and the plane below is $μ = 0.5$ and the plane below the $B$ block is frictionless. Assuming no slipping between pulley and string find acceleration of $4 kg$ block. Take $g = 10 m s^{- 2}$ .

Accepted Answer

Let us assume $4 kg$ block move in downward direction, $2 kg$ block moves in upward direction, string does not slip over pulley, it indicates pulley will move due to the torque provided by tension in the strings, therefore tension will be different in the strings. The blocks cover same magnitude of displacement, therefore their acceleration will have same magnitude, and it will be equal to the tangential acceleration of pulley.

(a). For block $A$ ,

$T_{1} - 2 g \sin (45 °) = 2 a \Rightarrow T_{1} - 10 \sqrt{2} = 2 a . . . (1)$

For block $B$ ,

$4 g \sin (45 °) - T_{2} = 4 a \Rightarrow 20 \sqrt{2} - T_{2} = 4 a . . . (2)$

For rotation of pulley, $(T_{2} \times r) - (T_{1} \times r) = I \times α$ for angular acceleration $α$ , $a = r α$ .

$T_{2} \times r - T_{1} \times r = I \times α \Rightarrow (T_{2} - T_{1}) \times 0.1 = 0.5 \times \frac{a}{0.1} \Rightarrow T_{2} - T_{1} = 50 a . . . (3)$ ,

Adding equation $(1), (2) & (3)$ ,

$- 10 \sqrt{2} + 20 \sqrt{2} = 56 a \Rightarrow 10 \sqrt{2} = 56 a \Rightarrow a = 0.25 m s^{- 2}$

Therefore, acceleration of block $4 kg$ is $0.25 m s^{- 2}$ .

(b). Now surface under block $A$ has friction, the frictional force will act in downward direction, the frictional force, $f = μ N_{1} = 0.5 \times 2 \times 10 \times \cos (45 °) N \Rightarrow f = 5 \sqrt{2} N$

For block $A$ ,

$T_{1} - 2 g \sin (45 °) - f = 2 a \Rightarrow T_{1} - 10 \sqrt{2} - 5 \sqrt{2} = 2 a \Rightarrow T_{1} - 15 \sqrt{2} = 2 a . . . (1)$

For block $B$ ,

$4 g \sin (45 °) - T_{2} = 4 a \Rightarrow 20 \sqrt{2} - T_{2} = 4 a . . . (2)$

For rotation of pulley,

$(T_{2} \times r) - (T_{1} \times r) = I \times α \Rightarrow (T_{2} - T_{1}) \times 0.1 = 0.5 \times \frac{a}{0.1} \Rightarrow T_{2} - T_{1} = 50 a . . . (3)$

Adding equations $(1), (2) & (3)$ , $- 15 \sqrt{2} + 20 \sqrt{2} = 56 a \Rightarrow 5 \sqrt{2} = 56 a \Rightarrow a = 0.125 m s^{- 2}$

Therefore, acceleration of block $4 kg$ is $0.125 m s^{- 2}$

Important Questions on Laws of Motion

Learn and Prepare for any exam you want !