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Question

Acceleration (a) of a body of radius R rolling down an inclined plane of inclination is given by?


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Solution

Step 1: Given data

K = Radius of gyration

R = Radius of the object

I = Inertia of the object

N = Normal force acting on the body

f = Frictional force

Step 2: Diagram

From the figure, mgsinθ&mgcosθ=components of force due to body's weight (mg)

Where, m=mass of the object, g= acceleration due to gravity, and θ = angle of the inclined plane

Step 3: Calculating acceleration

From the figure, we can see that the sine component of the weight of the object, mgsinθ acts towards left and the frictional force f acts opposite to the direction of motion. Since the body is rolling down, the resultant force,

ma=mgsinθ-f------(1)

Here, the body is rolling due to friction, therefore net torque is given by,

Torque = force ×perpendicular distance,

τ=fR

Also,

Torque = moment of inertia ×angular acceleration

τ=Iα

fR=mK²×αI=mK²-------(2)

K = radius of gyration

R = radius of the object

I = Inertia of the object

N = Normal force acting on the body

f = frictional force

In relation to linear and angular quantities, since there is no slipping

v=Rω

a=Rα----(3)

α=angular acceleration of the object

v=linear velocity of the object

ω=angular velocity of the object

Solving 1,2 &3 simultaneously,

From 2 Value of f=mK2αRsubstitute this value in 1

mgsinθ-mk2αR=magsinθ-K2αR=a-----(4)

From 3 α=aRsubstitute in 4

We get, a=gsinθ1+K2R2

Hence, the acceleration of a body rolling down an inclined plane is :a=gsinθ[1+K2R2]


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