Question

Q.5) (20 marks) Cylinder A is initially rotating at 1000 rpm in a clockwise direction; cylinder Bis rotating in a counterclockwise direction at 700 rpm. Cylinder A is brought into contact with cylinder B. Assuming no slipping, determine : Total initial momentum The final rpm of cylinder A Radius = 6 ft Wheel A: Solid cylinder, weight 128.8 lb Radius = 3 ft Wheel B: Solid cylinder, weight 64.4 lb Figure 5

Answer 1

Solution: Formula for moment of inertia of solid cylinder, $I=\frac{Mr^{2}}{2}$

Angular velocity, $\omega _{A}=\frac{2\pi N}{60}$

Wheel A

Moment of inertia,

angular vel, $\omega _{A}=104.71 rad/s$

Angular Momentum,

MA = LAWA = 2318.4 * 104.71 = 242759.664

Wheel B

Moment of inertia,

angular vel, $\omega _{B}=-73.30 rad/s$ (since angular velocity of B is in anti clockwise direction)

Angular Momentum, $M_{B}=I_{B}\omega _{B}=289.8*(-73.30)=-21242.34$

(a) Tota Initial momentum =$M_{A}+M_{B}$

=

ANS- 221517.324 is total initial angular momentum

(b) Final rpm of cylinder A- after the contact both the cylinder will rotate with common angular velocity $\omega$

$M_{A}+M_{B}=(I_{A}+I_{B})\omega$

$\frac{221517.324}{I_{A}+I_{B}}=\omega$

$\omega =84.9 rad/s$

ANS- final rpm of cylinder A is 84.9 rad/s