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Question
Discuss rolling on an inclined plane and arrive at the expression for acceleration.
Solution
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in the figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.
Rolling on an inclined plane
For transnational motion, mg sin θ is the supporting force and f is the opposing force, mg sin θ f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sin 0 cannot cause torque as it passes through it but the frictional force f can set torque of Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK2
we get,
Rf = mK2 `a/R`; f = ma `(K^2/R^2)`
Now equation becomes,
mg sin θ – ma `((K^2)/(R^2))` = ma
mg sin θ = ma + ma `(K^2/R^2)`
a `(1+K^2/R^2)` = g sin θ
After rewriting it for acceleration, we get,
a = `(gsinθ)/((1+K^2/R^2))`
We can also find the expression for the final velocity of the rolling object by using the third equation of motion for the inclined plane.
v2 = u2 + 2as. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = `h/sinθ`
`v^2 = 2(g sintheta)/((1 + K^2/R^2))(h/sin theta) = (2gh)/((1 + K^2/R^2))`
By taking square root,
v = `sqrt((2gh)/((1 + K^2/R^2)))`
The time taken for rolling down the incline could also be written from the first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,
t = `v/a`
t = `(sqrt((2gh)/((1 + K^2/R^2)))) (((1 + K^2/R^2))/(g sin theta))`
t = `sqrt((2h(1 + K^2/R^2))/(g sin^2theta))`
The equation suggests that for a given incline, the object with the least value of the radius of gyration K will reach the bottom of the incline first.
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