A turntable has a moment of inertia of 3.0 × 10−² kg·m² and spins freely on a frictionless bearing at 25 rev/min.A 0.60-kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10 m from the center.By what factor does the kinetic energy of the system change after the putty is dropped onto the turntable? A)0.91 B)1.0 C)0.83 D)1.5

Question

Quizplus · Accepted Answer

The initial kinetic energy of the system is purely rotational and given by:

K_rot = 1/2 * I * ω^2

where I is the moment of inertia, ω is the angular velocity in radians per second, and K_rot is the rotational kinetic energy.

Converting the initial angular velocity of the turntable to radians per second:

ω = (25 rev/min) * (2π rad/rev) * (1/60 min/s) = 2.62 rad/s

Substituting the given values:

K_rot,i = 1/2 * (3.0 × 10−8 kg·m2) * (2.62 rad/s)^2 = 41.9 mJ

After the putty is dropped on the turntable, the moment of inertia of the system changes to:

I_f = I + m*r^2

where m is the mass of the putty, r is the distance of the putty from the center, and I is the initial moment of inertia. Substituting the given values:

I_f = (3.0 × 10−8 kg·m2) + (0.60 kg)*(0.10 m)^2 = 3.06 × 10−8 kg·m2

Since the system is now a combination of rotational and translational motion, the total kinetic energy of the system is the sum of the rotational and translational kinetic energies:

K_total = K_rot + K_trans

where K_trans is the translational kinetic energy of the putty after it sticks to the turntable. The translational kinetic energy of an object is given by:

K_trans = 1/2 * m * v^2

where m is the mass of the object and v is its velocity.

The velocity of the putty after it sticks to the turntable is equal to the velocity of the turntable at the point where the putty sticks. This velocity can be found using the conservation of angular momentum:

I_i * ω_i = I_f * ω_f

where ω_i is the initial angular velocity of the turntable and ω_f is the final angular velocity of the turntable and putty together. Solving for ω_f:

ω_f = (I_i / I_f) * ω_i

Substituting the given values:

ω_f = (3.0 × 10−8 kg·m2 / 3.06 × 10−8 kg·m2) * 2.62 rad/s = 2.55 rad/s

The velocity of the putty is then:

v = r * ω_f = (0.10 m) * (2.55 rad/s) = 0.255 m/s

Substituting the given values:

K_trans = 1/2 * (0.60 kg) * (0.255 m/s)^2 = 4.37 mJ

Total kinetic energy after the putty is dropped:

K_total,f = K_rot,f + K_trans = 1/2 * I_f * ω_f^2 + K_trans

Substituting the given values:

K_total,f = 1/2 * (3.06 × 10−8 kg·m2) * (2.55 rad/s)^2 + 4.37 mJ = 42.3 mJ

The ratio of the final kinetic energy to the initial kinetic energy is:

K_total,f / K_total,i = 42.3 mJ / 41.9 mJ = 1.01

Therefore, the kinetic energy of the system increases by a factor of 1.01, or about 1.0. The closest answer choice is (C) 0.83, which is incorrect. The correct answer is (B) 1.0.

A Turntable Has a Moment of Inertia of 3