Question

1, A 0.30 m radius automobile tire accelerates from rest at a constant angular acceleration of 0.758 rad/s² for 2.23 s. What is its final angular speed in rad/s?

2. A turntable has a moment of inertia of 3.00 x 10^-2 kg-m² and spins freely on a frictionless bearing at 46 rev/min. A 1.11 kg ball of putty is dropped vertically onto the turntable and sticks at a point 0.100m from the center. What is the new rate of rotation of the system in rev/min?

Answer 1

Question 1

Radius of the automobile tire = R = 0.3 m

Initial angular speed of the tire = ₁ = 0 rad/s (At rest)

Final angular speed of the tire = ₂

Angular acceleration of the tire = = 0.758 rad/s²

Time period of acceleration = T = 2.23 sec

₂ = ₁ + T

₂ = 0 + (0.758)(2.23)

₂ = 1.69 rad/s

Final angular speed of the tire = 1.69 rad/s

Question 2

Moment of inertia of the turntable = I = 3 x 10^-2 kg.m²

Initial angular speed of the turn table = ₁ = 46 rev/min

Converting from rev/min to rad/s

₁ = 46 x (2/60) rad/s

₁ = 4.817 rad/s

Mass of the ball of putty = M = 1.11 kg

Distance of the point where the putty sticks from the center of the turntable = R = 0.1 m

New angular speed of the system = ₂

By conservation of angular momentum,

I₁ = (I + MR²)₂

(3x10^-2)(4.817) = (3x10^-2 + (1.11)(0.1)²)₂

₂ = 3.516 rad/s

Converting to rev/min,

₂ = 3.516 x (60/2) rev/min

₂ = 33.58 rev/min

New rate of rotation of the system = 33.58 rev/min