Arod of mass m = 116 g and length l = 49 cm can rotate about a hinge at its left end and is initially at rest. a putty ball of mass m = 14 g, moving with speed v= 6 m/s, strikes the rod at angle a = 37º a distance d=l/4 from the end and sticks to the rod after the collision.
(a) what is the total moment of inertia, 1, with respect to the hinge, of the rod-ball-system after the collision?
(b) enter an expression for the angular speed w of the system immediately after the collision, in terms of m, v, d, 0,
(c) calculate the rotational kinetic energy, in joules, of the system after the collision

a) Before:

I = 0.00928 kg*m^2

after:

I = 0.00949 kg*m^2

b) W = 65.25 rad/s

c) k = 20.2021 J

Explanation:

a)The moment of inertia before the collision is done by:

I =

where M is the mass and L is the length.

I =

I = 0.00928 kg*m^2

After the collision the moment of inertia is:

I =

the first part is the moment of inertia before the collision but now we plus the mass of the putty ball ( and the distance R where the putty ball strickes.

I =

I = 0.00949 kg*m^2

b)

For answer this we will use the law of the conservation of the angular momentum:

Adittionally, the angular momentum L could be calculated by:

L = MDV

where M is the mass, V is the velocity and D is the lever arm, or:

L = IW

where I is the moment of inertia and W is the angular velocity.

Following the equations, now we have:

where is the mass of the putty ball, is the velocity of the putty ball, and is the moment of inertia after the collition

replacing the data:

solving for W, we get:

W = 65.25 rad/s

c)

The kinetic energy is calculated by the next equation:

K =

replacing the data:

k =

k = 20.2021 J

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