Physic's help?

Question

A bowling ball of mass 7.1 kg and radius 9.0 cm rolls without slipping down a lane at 4.9 m/s. Calculate its total kinetic energy.
J

Answer 1

The ball will possess both linear and rotational kinetic energy (KE), and so the total KE will be the sum of these parts.

The linear KE is given by (1/2)*m*v^2, where m is the mass of the ball, and v is its speed.

For the rotational KE, we use the analogous equation (1/2)*I*w^2, where I is the moment of inertia of the ball, and w is the angular velocity. Now, the moment of inertia of a body depends on its shape and the axis about which it is rotating, and for a uniform, solid sphere rotating about its centre, I = (2/5)*m*r^2, where r is the radius (see http://en.wikipedia.org/wiki/List_of_moments_of_inertia ).

In the question, you say that the ball is not slipping. This means that in one rotation of the ball, it has moved a linear distance equal to its circumference. The appropriate result from this is that the linear velocity v and the angular velocity w are related through v = r*w. Hence, the rotational KE becomes (1/2)*I*(v/r)^2.

Now, adding together the linear and rotational KEs, and putting in the expression for the moment of inertia I, the total kinetic energy E is given by

E = (1/2)mv^2 + (1/2)((2/5)mr^2)(v/r)^2

E = (1/2)mv^2 [ 1 + (2/5)]

E = (7/10)mv^2.

Putting in the values for m and v, we get

E = (7/10)*(7.1 kg)*(4.9 m/s)^2 = 119.3297 J.

So, the answer is about 120 J, to the same degree of accuracy as the values given in the question.

Answer 2

It's total kinetic energy is the kinetic energy of the center of mass plus it's rotational kinetic energy.

Answer 3

Use the formula :: K.E. = 1/2 mv^2
for v use the formula of sphere