Question

A thin, 100 g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the speed of a point on the rim

Answer 1

Answer:

v=2.45 m/s

Explanation:

The rotational kinetic energy is calculated with:

$K=\frac{I\omega^2}{2}$

The moment of inertia of a disk is:

$I=\frac{mr^2}{2}$

A point on the rim of a disk will have a speed of:

$v=r\omega$

Putting all together:

$v=r\sqrt{\frac{2K}{I}}=r\sqrt{\frac{2(2K)}{mr^2}}=2\sqrt{\frac{K}{m}}$

Which for our values is:

$v=2\sqrt{\frac{0.15J}{0.1Kg}}=2.45m/s$

Answer 2

Given Information:

kinetic energy = E = 0.15 J  

mass = m = 100g = 0.1 kg

diameter = d = 8.0 cm = 0.08 m

Required Information:

Speed = v = ?

Answer:

v = 2.44 m/s

Explanation:

First convert diameter into radius

r = d/2 = 0.08/2 = 0.04 m

The moment of inertia of disc is given by

I = 0.5mr²

Where m is the mass and r is the radius of the disk

I = 0.5(0.1)(0.04)²

I = 0.00008 kg.m²

and since we are dealing with rotational motion then rotational kinetic energy is given by

E  =  0.5 I ω²

we have to separate ω

ω² = E/0.5 I

ω = √E/0.5 I

ω = √0.15/0.5(0.00008)

ω = 61.23 rad/sec

Finally we know that speed is given as

v = ω r

v = 61.23 (0.04)

v = 2.44 m/s

Therefore, the speed of a point on the rim  is 2.44 m/s