a grinding wheel of mass 5.0kg and diameter 0.4 m is rotating with an angular speed of 2 rev S-1,calculate the torque which will increase its angular speed to 8 revolution per second in 2s.
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Hello dear,
◆ Answer-
τ = 251.33 Nm
◆ Explanation-
# Given-
m = 5 kg
r = 4/2 = 2 m
f1 = 2 rps
f2 = 8 rps
t = 2 s
# Solution-
Rotational acceleration is-
α = (ω2-ω1) / t
α = 2π(f2-f1) / t
α = 2π(8-2) / 2
α = 4π rad/s^2
Moment of inertia is -
I = mr^2
I = 5 × 2^2
I = 20 kgm^2
Rotational torque is calculated by-
τ = Iα
τ = 20 × 4π
τ = 251.33 Nm
Therefore, torque required is 251.33 Nm.
Hope this helps you...
◆ Answer-
τ = 251.33 Nm
◆ Explanation-
# Given-
m = 5 kg
r = 4/2 = 2 m
f1 = 2 rps
f2 = 8 rps
t = 2 s
# Solution-
Rotational acceleration is-
α = (ω2-ω1) / t
α = 2π(f2-f1) / t
α = 2π(8-2) / 2
α = 4π rad/s^2
Moment of inertia is -
I = mr^2
I = 5 × 2^2
I = 20 kgm^2
Rotational torque is calculated by-
τ = Iα
τ = 20 × 4π
τ = 251.33 Nm
Therefore, torque required is 251.33 Nm.
Hope this helps you...
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