Name: The Relation Between Torque and Angular Acceleration
Uploaded: 2019-07-19
Duration: 4 min 7 s
Description: In translational motion, force is related to acceleration. Likewise, in rotational motion, torque is related to angular acceleration! Learn how the role of m...

Question

A uniform cylinder of radius

12 cm

and mass

25 kg

is mounted to rotate freely about a horizontal axis that is parallel to and

5.0 cm

from the central longitudinal axis of the cylinder.
(a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Accepted Answer

Given the radius of the cylinder, $r = 12 cm = 0.12 m$ , mass of the cylinder, $m = 25 kg$ and the separation between the rotational axis and the centre of the cylinder, $h = 5 \times 10^{- 2} m$ ,

$(a)$ Now, inertia of rotation about the axis of rotation is given by,

$I = I_{cm} + m h^{2} I = \frac{1}{2} m r^{2} + m h^{2} = \frac{1}{2} \times 25 \times 0.12 \times 0.12 + 25 \times 0.05 \times 0.05 = 0.18 + 0.0625 = 0.2425 kg m^{2}$

$(b)$ From the conservation of total energy, we know that

$\frac{1}{2} I (ω^{2} - {ω^{2}}_{i}) = u_{i} - u_{f}$ , here change in rotational kinetic energy is, $\frac{1}{2} I (ω^{2} - {ω^{2}}_{i})$ .

$\Rightarrow u_{i} - u_{f} = m g h = \frac{1}{2} \times 0.2425 \times ω^{2} = 25 \times 9.8 \times 0.05 ω = \sqrt{\frac{2 \times 25 \times 9.8 \times 0.05}{0.2425}} \approx 10 rad s^{- 1}$

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