Question 666: Centre of Mass and Rotational Motion

Question

The instantaneous angular position of a point on a rotating wheel is given by the equation $$	heta \left( t ight) = 2{t^3} - 6{t^2}$$
The torque on the wheel becomes zero at

Accepted Answer

When angular acceleration (a) is zero then
torque on the wheel becomes zero.
$$	heta \left( t ight) = 2{t^3} - 6{t^2}$$
$$ \Rightarrow {{d	heta } \over {dt}} = 6{t^2} - 12t$$
$$ \Rightarrow \alpha  = {{{d^2}	heta } \over {d{t^2}}} = 12t - 12 = 0$$
$$ 	herefore $$ t = 1 sec.

Answer

t = 1 s

Answer

t = 0.5 s

Answer

t = 0.25 s

Answer

t = 2 s

Centre of Mass and Rotational Motion

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta \left( t \right) = 2{t^3} - 6{t^2}$ The torque on the wheel becomes zero at

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