 A pulley in a well with a diameter of 1200 cm initially rotates at 1.3 rev/s and then receives a constant angular acceleration of 3.12 rad/s2 . What is the tangential velocity of a belt mounted on said pulley? After 1 second, what is the tangential acceleration of the belt?

1. Calculate the initial angular velocity in radians per second:
\omega_0 = 1.3 \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 2.6\pi \, \text{rad/s}

2. Determine the final angular velocity after 1 second using the equation for angular velocity with constant acceleration:
\omega = \omega_0 + \alpha t
where \alpha = 3.12 \, \text{rad/s}^2 and t = 1 \, \text{s}

3. Calculate \omega :
\omega = 2.6\pi + 3.12 \times 1 = 2.6\pi + 3.12

4. Calculate the radius of the pulley:
r = \frac{1200 \, \text{cm}}{2} = 600 \, \text{cm} = 6 \, \text{m}

5. Find the tangential velocity v at t = 1 \, \text{s} :
v = \omega \times r
v = (2.6\pi + 3.12) \times 6

6. Simplify to find v :
v = (8.168 + 3.12) \times 6 = 11.288 \times 6 = 67.728 \, \text{m/s}

7. Calculate the tangential acceleration a_t , which is constant:
a_t = \alpha \times r = 3.12 \times 6 = 18.72 \, \text{m/s}^2

8. Final answers:
Tangential velocity after 1 second: 67.728 \, \text{m/s}
Tangential acceleration: 18.72 \, \text{m/s}^2