Question

Wheel A of radius rA = 10.0 cm is coupled by a belt B to wheel C of radius rC = 25.0 cm, as shown in Fig.
Wheel A increases its angular speed from rest at a uniform rate of 1.60 rad/s2, Determine the time for wheel C to reach a rotational speed of 100 rev/min, assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the rims of the two wheels must be equal.)

Answer 1

$t\approx16.36\ s$ is the time taken for the wheel C  to reach a speed of 100 rpm.

Explanation:

Given:

• radius of wheel A, $r_a=0.1\ m$

• radius of wheel C, $r_c=0.25\ m$

• angular acceleration of wheel A, $\alpha_a=1.6\ rad.s^{-2}$

• final rotational speed of wheel C, $N_c=100\ rpm$

So, linear velocity of wheel C at final condition:

$v_c=\frac{2\pi\times r_c\times N}{60}$

$v_c=\frac{2\pi\times 0.25\times 100}{60}$

$v_c=2.618\ m.s^{-1}$

According to the condition of no slip:

$v_c=v_a$

$v_c=r_a.\omega_a$

$\omega_c=\frac{2.618}{0.1}$

$\omega_c=26.18\ rad.s^{-1}$ is the final velocity for wheel A.

Now time taken to reach this angular velocity:

using equation of motion

$\omega_c=\omega_{ci}+\alpha_a.t$

$26.18=0+1.6\times t$

$t\approx16.36\ s$

TOPIC: angular velocity, tangential velocity, equation of motion

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