Question

A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angular velocity ω of the wheel at the bottom of the incline?
Express your answer in radians per second.

Answer 1

The angular velocity of the wheel depends on the mass, radius and the

mode of rotation of the wheel (with or without slipping).

• The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/sec

Reasons:

The given parameters are;

Radius of the wheel, r = 2.0 m

Height of the incline, h = 8.0 m

Required:

Angular velocity of the wheel at the bottom of the incline.

Solution:

The potential energy of the wheel at the top of the hill, P.E. = m·g·h

$Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}$

Where;

v = The translational velocity of the wheel = ω·r

I = The moment of inertia of the wheel = m·r²

Therefore'

$Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}$

At the bottom of the hill, the potential energy is converted to kinetic energy

Therefore;

P.E. = Sum of K.E.

m·g·h = m·r²·ω²

g·h = r²·ω²

$\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}$

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}$

• The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/sec

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Answer 2

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

$\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}$

$\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}$

$\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}$

$\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }$

$\mathbf{\omega^2=\dfrac{39.24 }{2}}$

$\mathbf{\omega=\sqrt{19.62 } \ rad/sec}$

$\mathbf{\omega=4.429 \ rad/sec}$

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

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