In a previous more detailed post (June 28, 2019) I examined the rotational dynamics of an object that i) rolls without slipping and ii) rolls down an inclined plane. E.g.

So that the linear distance covered, S, for an object of radius R, is related to the angular rate covered (dq /dt ) by

dS/ dt - R (dq /dt) = 0

*const*. = 0

In terms of case (ii) we saw:

For which the moment of inertia of the rolling disc: I = ½ m R ^{2}

had to be taken into account. And we have the differential equation:

^{2}+ ½ I q’

^{2}

The last equation actually provides the clue for dealing with the application of energy conservation to rolling objects in a more basic framework, i.e. without calculus. That is, one needs to determine the kinetic energy (K) of rolling motion which combines rotation and translation. Thus we have:

K = ½ Iw_{ }^{2 } + ½M v_{ }^{2}

In the preceding, I is the moment of inertia about the center of the rolling object. The equation for K can actually be simplified by using the fact that the linear and angular speeds are related, as we saw in Part 1:

v = r (q/ t) = r w.

Therefore we can rewrite K with w= v/r :

K = ½ I (v/r )^{2 } + ½M v_{ }^{2}

= ½M v_{ }^{2} (1 + I/M r_{ }^{2} )

Since I = (const) (M r_{ }^{2}) the last term in the preceding equation is a constant that depends on the shape and mass distribution of the rolling object.

*Example Problem*:

A 1.20 kg disk with a radius of 10.0 cm rolls without slipping. If the linear speed is 1.41 m/s find: a) the translational kinetic energy, b) the rotational kinetic energy and c) the total kinetic energy of the disk.

* Solution*:

Because the disk rolls without slipping the angular speed and the linear speed are related by: v = r w

a) the translational kinetic energy,

We have: K = ½M v_{ }^{2}

^{= }½ (1.20 kg) (1.41 m/s)^{2} = 1.19 J

b) the rotational kinetic energy:

Using algebra we can write:

K = ½ Iw_{ }^{2 }= ½ (½ M r_{ }^{2} ) (v/r )^{2} = ½ (½ Mv_{ }^{2} )

But we saw from part (a): ½ Mv_{ }^{2} = 1.19 J

So: K = ½ Iw_{ }^{2 }= ½ (1.19 J) = 0.595 J

c) The total kinetic energy of the disk

This is simply the sum of parts (a) and (b):

K = ½ I (v/r )^{2 } + ½M v_{ }^{2}= 1.19 J + 0.595 J = 1.79 J

The case of rolling motion down an incline is now considered based on the sketch below:

In this case we have an object of mass m, radius r and moment of inertia I rolling down a ramp from a height h and want to find the speed at the bottom. The principle of energy conservation is again the best approach to solving this problem. At the top of ramp at height h we can write the sum of kinetic and potential energies as:

K _{i} + V _{i}
= 0 + mgh = mgh

Then at the bottom we will have:

K _{f} + V _{f}
= _{ }^{2} (1 + I/m r_{ }^{2} ) + 0

Since: K _{i} + V _{i} = K _{f} + V _{f}

Then:

mgh = ½ m v_{ }^{2} (1 + I/m r_{ }^{2} )

Solving for v:

v = Ö2gh/ (1 + I/m r_{ }^{2} )

For the special case of I = 0:

v = Ö 2gh

__Suggested Problems:__

1) A 1.0 kg hollow sphere with a radius of 10.0 cm rolls without slipping. If the linear speed is p/2 m/s find: a) the translational kinetic energy, b) the rotational kinetic energy and c) the total kinetic energy of the disk. (The moment of inertia for a hollow sphere is: I = 2/3 m R ^{2} )

2) A 0.5 kg solid sphere with a radius of 5.0 cm rolls without slipping down an incline of height h = 0.5m. Find: a) the initial total energy (kinetic and potential), and b) the final total energy, and c) the velocity at the bottom of the incline. ((The moment of inertia for a solid sphere is: I = 2/5 m R ^{2} )

3) A 1.0 kg solid ball rolling on a horizontal surface at 20 m/s arrives at the bottom of an inclined plane which makes an angle of 30 degrees with the horizontal. Find: a) The total kinetic energy of the ball when it has just arrived at the bottom of the incline, b) The distance up the incline the ball will continue to roll. (Ignore frictional forces). *See sketch below:*

4) A wheel of radius 0.5m and mass 5.0 kg rolls down a smooth incline 15 m high. Find the wheel's linear speed when it is at the bottom of the incline. (Take the moment of inertia as I = m R ^{2} )

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