Question

# A 2.15-kg, 16.0-cm radius, high-end turntable is rotating freely at 33.3 rpm when a naughty child...

A 2.15-kg, 16.0-cm radius, high-end turntable is rotating freely at 33.3 rpm when a naughty child drops 11 g of chewing gum onto it 14.0 cm from the rotation axis.

1)

Assuming that the gum sticks where it lands, and that the turntable can be modeled as a solid, uniform disk, what is the new angular speed of the turntable?

angular velocity w1 = 33.3 rpm
moment of inertia of turntable I1 = 0.5*M*R^2
Mass of turntable = M = 2.15 kg
Radius of turntable R = 16.0 cm = 0.16 m
Moment of inertia of turntable(I1) = 0.5*M*R^2

I1 = 0.5*2.15*(0.16)^2

I1 = 0.02752 kg*m^2
Total moment of inertia of turntable+ chewing gum (I2) = I1 + m*r^2
Or I2 = 0.02752 + m*r^2

here, m = mass of gum = 11 gm = 0.011 kg

r = distance of gum from center = 14.0 cm = 0.14 m

So, I2 = 0.02752 + 0.011*0.14^2

I2 = 0.02774 kg*m^2
Let w2 = final angular velocity = ??
By conservation of angular momentum,
Li = Lf
I1 * w1 = I2 * w2
Or 0.02752*33.3 = 0.02774*w2
Or w2 = 0.02752*33.3/0.02774
w2 = 33.04 rpm = new angular speed of the turntable

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