Question

A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?

Answer 1

The moment of inertia of the wheel is 4.27 kg.m²

The kinematics equation explains the variables associated and related of motion.

From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:

$\mathbf{y = ut + \dfrac{1}{2}at^2}$

$\mathbf{y = (0)t + \dfrac{1}{2}at^2}$

$\mathbf{y = \dfrac{1}{2}at^2}$

Making acceleration (a) the subject, we have:

$\mathbf{a = \dfrac{2y}{t^2}}$

where;

• y = 1.5 m
• t = 2.0 s

$\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}$

a = 0.75 m/s²

The angular acceleration of the wheel can be estimated by the formula:

$\mathbf{\alpha = \dfrac{a}{r}}$

$\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}$

$\mathbf{\alpha = 1.875 \ rad/s^2}$

Finally, the torque acting on the wheel is:

$\mathbf{\tau = I \alpha}$

$\mathbf{Tr = I \alpha}$

where;

• T = tension
• r = radius

• I = moment of inertia
• ∝ = angular acceleration

∴

$\mathbf{I =\dfrac{T\times r}{\alpha} }$

$\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }$

I = 4.27 kg.m²

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