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Question: The descending pulley shown in figure has a radius \(20cm\) and moment of inertia \(0.20kg{m^2}\). T...

The descending pulley shown in figure has a radius 20cm20cm and moment of inertia 0.20kgm20.20kg{m^2}. The fixed pulley is light and horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0kg1.0kg.

Explanation

Solution

Understand the forces acting on the block and the movable pulley and the directions of all these forces. To find out the value of acceleration, equate these forces as per the direction of the forces.

Complete step by step solution:
In the figure, there are two pulleys, light pulley A & descending pulley B.
Descending pulley B have radius 20cm20cm and moment of inertia 0.20kgm20.20kg{m^2}
Mass of the block is 1.0kg1.0kg as shown in the figure.

The acceleration of the block=aa
The acceleration of the descending pulley =a′a'
Tension acting on the left side of the movable pulley= tension acting on the block= TT
Tension acting on the right side of the movable pulley= T′T'
Mass of the movable pulley= mm
From the figure we can understand that, the acceleration of the block= double the acceleration of the descending pulley’
That is, a=2a′a = 2a'
⇒a′=a2\Rightarrow a' = \dfrac{a}{2}………………… (1)
When we consider the block only,

The forces acting on the block are: Gravitational force, normal force, tension, and also there is an acceleration in the direction of the string that is also present.
That is, T=aT = a ………….. (2)

When we consider the movable pulley only,

The forces acting on the movable pulley are: Gravitational force, tension on the right side, tension on the left side and also there is an acceleration in downward direction is also present.
That is, mg−T−T′=ma′mg - T - T' = ma'…………………….. (3)
The torque acting will be about the centre of the movable pulley will be,
τ=(T′−T)×R=I×α\tau = (T' - T) \times R = I \times \alpha ……………….. (4)
Here α\alpha is the angular acceleration and the value given by,
α=a′R=a2R\alpha = \dfrac{{a'}}{R} = \dfrac{a}{{2R}}
The value of II is given by,
I=mR22I = \dfrac{{m{R^2}}}{2}
Applying these values to the equation (4) we get,
τ=(T′−T)×R=mR22×a2R\tau = (T' - T) \times R = \dfrac{{m{R^2}}}{2} \times \dfrac{a}{{2R}}
⇒\Rightarrow ⇒(T′−T)=m4a \Rightarrow (T' - T) = \dfrac{m}{4}a………………….. (5)
It given in the question that, moment of inertia 0.20kgm20.20kg{m^2}
That is, I=mR22=0.2I = \dfrac{{m{R^2}}}{2} = 0.2
From this equation we get,
m=0.2×2(0.2)2m = \dfrac{{0.2 \times 2}}{{{{(0.2)}^2}}}
⇒m=10kg\Rightarrow m = 10kg
Applying this value in the equation (5) we get,
⇒(T′−T)=m4a\Rightarrow (T' - T) = \dfrac{m}{4}a
⇒(T′−T)=104a=52a\Rightarrow (T' - T) = \dfrac{{10}}{4}a = \dfrac{5}{2}a
⇒T′=52a+a=72a\Rightarrow T' = \dfrac{5}{2}a + a = \dfrac{7}{2}a

Applying values in equation (3)
mg−T−T′=ma′mg - T - T' = ma'
⇒(10×9.8)−a−7a2=10a2\Rightarrow (10 \times 9.8) - a - \dfrac{{7a}}{2} = \dfrac{{10a}}{2}
⇒98−9a2=5a\Rightarrow 98 - \dfrac{{9a}}{2} = 5a
∴a=10.316ms−2\therefore a = 10.316m{s^{ - 2}}

Final answer is, the acceleration of the block, a=10.316ms−2a = 10.316m{s^{ - 2}}.

Note: A pulley is a wheel on an axle or shaft that is designed to support movement and change of direction of a taut cable or belt, or transfer of power between the shaft and cable or belt. The movable pulley reduces the effort required to pull a mass by factor of 2×2 \times number of movable pulleys.