Question

Two gears are both connected to the same belt. The smaller of the two gears has a radius of 4ft & the larger of the two gears has a radius of 7 ft. The larger gear is turning at a rate of 50 revolutions per minute. What is the angular velocity of the larger gear in radians/second? What is the angular velocity of the smaller gear in radians/second?

Answer 1

Answer:

$\omega_1 = \frac{5Rad}{3s}$ and $\omega_2 = \frac{35Rad}{12s}$

Step-by-step explanation:

Given that the larger gear is turning at the rate 50rpm. The angular velocity, $\omega_1$, of the larger gear can be calculated as follows:

$\omega_1 = \frac{50rev}{m}$ to convert that to secs, we know that 1 rev = 2πRad and 60secs make a min, so we have:

$\omega_1 = \frac{50rev}{m} \times \frac{2\pi Rad}{1rev} \times \frac{1m}{60s}$

That gives

$\omega_1 = \frac{100Rad}{60s}$

Which gives:

$\omega_1 = \frac{5Rad}{3s}$

This is the angular velocity of the larger gear in rads/sec.

In order to get the angular velocity of the smaller gear, we know from the question that both gear are connected to the same belt. This implies that they have the same linear velocity. Thus let us find the linear velocity of the larger gear. This is given by

$v = r_1 \times \omega_1$ where v stands for linear velocity and $r_1$ for radius of larger velocity.

We are given the radius of the larger gear as 7ft so we have

$v = 7ft \times \frac{5Rad}{3s} = \frac{35ft}{3s}$

To get the angular velocity of the smaller gear, we use

$v = r_2 \times \omega_2$ where $r_2$ is the radius of the smaller gear which is given as 4.

Since linear velocity is the same, we have

$\omega_2 = \frac{35ft}{4 \times 3s}$

$\omega_2 = \frac{35Rad}{12s}$