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?

Respuesta :

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:

[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]

[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]

[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]

Making acceleration (a) the subject, we have:

[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]

where;

  • y = 1.5 m
  • t = 2.0 s

[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]

a = 0.75 m/s²

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

[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]

[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]

[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]

Finally, the torque acting on the wheel is:

[tex]\mathbf{\tau = I \alpha}[/tex]

[tex]\mathbf{Tr = I \alpha}[/tex]

where;

  • T = tension
  • r = radius
  • I = moment of inertia
  • ∝ = angular acceleration

[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]

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

I = 4.27 kg.m²

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