Answer:
(a) ∝ = 18.84 rad/s^2
(b) 0.297 N m
Explanation:
m = 0.35 kg
r = 0.3 m
fo = 0 rps
f = 24 rps
t = 8 s
(A)
ωo = 2 x π x fo = 0
ω = 2 π f = 2 x 3.14 x 24 = 150.72 rad/s
Use first equation of motion for rotational motion
ω = ωo + ∝ t
Where, ∝ is the rotational acceleration
150.72 = 0 + 8 ∝
∝ = 18.84 rad/s^2
(B) Torque = Moment of inertia x rotational acceleration
Moment of inertia of turn table = 0.5 x mass x radius²
I = 0.5 x 0.35 x 0.3 x 0.3 = 0.01575 kg m²
Torque = 0.01575 x 18.84 = 0.297 N m
The rotational acceleration of the turn table is 18.85 rad/s².
The torque required to cause the acceleration is 0.298 Nm.
The rotational acceleration of the turn table is calculated as follows;
[tex]\alpha = \frac{\Delta \omega}{t} = \frac{24(2\pi) - 0}{8} = 18.85\ rad/s^2[/tex]
The moment of inertia of the turn table is calculated as follows;
[tex]I = \frac{1}{2} mr^2\\\\I = \frac{1}{2} \times 0.35 \times 0.3^2\\\\I = 0.0158 \ kgm^2[/tex]
The torque required to cause the acceleration is calculated as follows;
[tex]\tau = I \alpha\\\\\tau = 0.0158 \times 18.85\\\\\tau = 0.298 \ Nm[/tex]
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