Question

Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle will it have turned after time 2t?

Answer 1

Answer:

After time 2t, the wheel will turn through an angle of 100 rad.

Explanation:

Given;

initial velocity of the wheel, $\omega_i = 0$

time of motion, t = t

angular distance, θ = 25 rad

The constant angular acceleration is calculated as;

$\theta = ut + \frac{1}{2}at^2\\\\\theta = 0 + \frac{1}{2}at^2\\\\2\theta = at^2\\\\a = \frac{2\theta }{t^2} \\\\a = \frac{2\ \times \ 25 }{t^2}\\\\a = \frac{50}{t^2} \\\\when \ t = 2t\\\\\theta = \frac{1}{2}at^2\\\\\theta = \frac{1}{2}(\frac{50}{t^2})(2t)^2\\\\\theta =\frac{1}{2}(\frac{50}{t^2})(4t^2)\\\\\theta = 100 \ rad$

Therefore, after time 2t, the wheel will turn through an angle of 100 rad.

Answer 2

The angle of the wheel would have turned after time 2t is equal to 100 rad.

Given the following data:

• Initial angular velocity = 0 rad/s (since the wheel starts from rest).

• Time = t seconds

• Angle = 25 rad

To determine the angle of the wheel would have turned after time 2t:

First of all, we would determine the angular acceleration of the wheel by using the second equation of rotational motion.

$\theta = ut + \frac{1}{2} at^2\\\\25 =0(t) + \frac{1}{2} at^2\\\\25=\frac{1}{2} at^2\\\\50=at^2\\\\a=\frac{50}{t^2} \;rad/s^2$

Next, we would determine the angle the wheel would have turned after time 2t:

$\theta = ut + \frac{1}{2} at^2\\\\\theta = 0(2t) + \frac{1}{2} (\frac{50}{t^2} )(2t)^2\\\\\theta = \frac{25}{t^2} \times 4t^2\\\\\theta = 25\times 4\\\\\theta = 100\;rads$

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