Finding New Rate of Rotation for a Student on a Rotating Platform with Dumbbells

[chess10771]

Homework Statement

A student stands on a platform that is free to rotate and holds two dumbbells, each at a distance of 65 cm from his central axis. Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.95 rev/s. The student on the platform then pulls the dumbbells in close to his chest so that they are each 22 cm from his central axis. Each dumbbell has a mass of 5.00 kg and the rotational inertia of the student, platform, and dumbbells is initially 7.40 kg · m2. Model each arm as a uniform rod of mass 3.00 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 22 cm. What is his new rate of rotation?

Homework Equations

Newtons 2nd law?

The Attempt at a Solution

Not really sure how to go about solving, Please show all work if possible, thanks in advance.

 

[TVP45]

I'd guess you're in a freshman physics class this semester, probably about Chapter 8, +/- 1 Chapter. So, what are you studying that looks anything like this? Skim through the text and look for a picture or diagram that suggests anything close to this.

 

[kerimek]

I would approach this problem by first identifying the relevant equations and principles that can be applied. In this case, the relevant concepts are rotational motion and conservation of angular momentum.

The initial rate of rotation, or angular velocity, of the system can be calculated using the formula ω = v/r, where ω is the angular velocity, v is the linear velocity, and r is the distance from the central axis. In this case, the initial angular velocity can be calculated as 0.95 rev/s = 6.0 m/s / 0.65 m.

Next, we can use the principle of conservation of angular momentum to calculate the new rate of rotation. This principle states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the initial angular momentum of the system is equal to the final angular momentum, since there is no external torque acting on the system.

The initial angular momentum can be calculated as Iω, where I is the rotational inertia and ω is the angular velocity. Substituting the given values, we get an initial angular momentum of 7.40 kg · m^2 * 0.95 rev/s = 7.03 kg · m^2/s.

To find the final angular momentum, we need to use the equation L = Iω, where L is the angular momentum, I is the rotational inertia, and ω is the final angular velocity. Since the rotational inertia has changed due to the student pulling the dumbbells closer to his chest, we need to calculate the new rotational inertia. This can be done by using the parallel axis theorem, which states that the rotational inertia of a body about an axis parallel to its center of mass is equal to the rotational inertia about the center of mass plus the mass of the body times the square of the distance between the two axes.

Applying this to our problem, the new rotational inertia can be calculated as I' = I + md^2, where I is the initial rotational inertia, m is the mass of the dumbbell (5.00 kg), and d is the distance between the initial and final positions of the dumbbell (43 cm). Substituting the values, we get I' = 7.40 kg · m^2 + (5.00 kg)(0.43 m)^2 = 7.93 kg · m^2.

Now, we can use the conservation