How to Calculate Angular Acceleration and Kinetic Energy?

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SUMMARY

This discussion focuses on calculating angular acceleration and kinetic energy in various scenarios involving rotational motion. The first problem involves a turntable rotating at 32 1/3 revolutions per minute (rev/min) that comes to rest in 29 seconds, yielding an angular acceleration of -0.018582 s-2 and a total of 8.03 revolutions. The second problem calculates the kinetic energy of a solid ball with a mass of 1.6 kg and a diameter of 20 cm rotating at 64 rev/min, resulting in a kinetic energy of 0.575 J. Lastly, the discussion addresses the speed of a 1300-kg car falling from rest, which is calculated to be 8.89 m/s using principles of energy conservation.

PREREQUISITES
  • Understanding of angular motion and acceleration
  • Familiarity with the moment of inertia (I) and its calculation
  • Knowledge of kinetic energy formulas in rotational dynamics
  • Basic principles of energy conservation in mechanics
NEXT STEPS
  • Study the derivation of angular acceleration formulas in rotational dynamics
  • Learn about the conservation of mechanical energy in falling objects
  • Explore the relationship between linear and angular velocity
  • Investigate the effects of moment of inertia on rotational motion
USEFUL FOR

Students in physics, mechanical engineers, and anyone interested in understanding the principles of rotational motion and energy calculations.

Gtseviper
Please help

When a turntable rotating at 32 1/3 rev/min is shut off, it comes to rest in 29 s.

a assuming constant angular accelerationn, find the angular acceleration.
(my work) W-Wo/t=a and I got -.12

b number of revolutions
(my work) phida-initial phida= Wot + 1/2at^2 and I got 8.03 rev

2 A solid ball of mass 1.6kg and diameter 20 cm is rotating about its diameter at 64rev/min

a What is its KE.
(my work) K= 1/2 Iw^2, I=2/5 MR^2 I= 2.56 x10^-2 kg m^2 K= 5.75 x 10^-1 J

b addition 2 J is added, new angular speed.
(my work) 2.57 J is new K, W^2= 2K/I W= 1.36 x 10^2 rev/min

3 A 1300-kg car is being unloaded by a winch. At the moment shown in Figure 9-46, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 300 kg · m2 and that of the pulley is 2 kg · m2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

and I got 8.89 by using mgh= 1/2 Iw (V/rw)^2+ 1/2 Ip (V/rp)^2 mv^2
 

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For question 1, the turntable. 32 1/3 revs per minute gives ω0 = 0.5389 s-1.
The angular acceleration (assuming constant) gives α = (0-ω0)/t = - 0.5389 / 29 = 0.018582 s.
Can you find the correct number of revolutions now? Keep in mind that one revolution is 2π
 

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