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# Rotational dynamics

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A turn-table on a frictionless vertical axis has a radius of 1.5 m and a moment of rotational inertia of 150 kgm^2. It is at rest until a 75 kg person who had been standing still (and who can be treated as a point), begins to walk around the rim at a speed of 2.0 m/s with respect to the turn-table.
a) what is the speed of the person with respect to the Earth?
b)How much work did the person do in bringing the turn-table up to speed?

https://brainmass.com/physics/angular-momentum/rotational-angular-speed-turn-table-5757

## SOLUTION This solution is FREE courtesy of BrainMass!

a) what is the speed of the person with respect to the Earth?
Moment of inertia of turntable= 150 Kg m^2 =It

Mass of the person= 75 Kg
"Moment of inertia of the person with respect to the vertical axis=
M r ^2 =" 168.75 Kg m^2 =Ip

Let the person be moving with angular velocity w p with respect to the earth
Let the turntable be moving with angular velocity w t with respect to the earth

w p= Vp/r where Vp is the velocity of the person with respect to the earth
w t= Vt/r where Vt is the velocity of the rim of the turntable with respect to the earth

Since the angular momentum is conserved and since the turntable and the person were initially
at rest

Ip w p + It w t=0
Vector equation
or 168.75 * Vp/r +150 * Vt/r =0

or 168.75* Vp + 150 * Vt=0 (since r is common, r is not zero and RHS= 0 r can be removed from the equation)

Thus Vp and Vt are in opposite direction.
If we use Vp and Vt to denote the magnitude of velocities and treat the direction of Vp as positive and
the direction of Vt as negative

The equation becomes
Scalar equation
168.75* Vp - 150 * Vt=0

or Vt=168.75/150= 1.125 Vp --------(1)

We also know from the data in the problem that the person is moving at a speed of 2.0 m/s with respect to the turn-table.

Thus Vp- Vt =2 Vector equation
If we use Vp and Vt to denote the magnitude of velocities and treat the direction of Vp as positive and
the direction of Vt as negative
Scalar equation
Vp+ Vt =2 --------(2)

Solving simultaneously Equation 1 and 2

Vp+1.125 Vp =2
or Vp= 0.94 m/s
Vt=1.125 Vp = 1.06 m/s
The speed of the person with respect to the Earth= 0.94 m/s

b)How much work did the person do in bringing the turn-table up to speed?

Rotational Kinetic energy = 1/2 I w ^2

Rotational Kinetic energyof person = 1/2 Ip w p^2= 33.49 Joules
Rotational Kinetic energyof turntable = 1/2 It w t^2= 37.81 Joules
Total= 71.30 Joules

This is the total amount of work done by the person= 71.30 Joules
as initially both the person and turntable
were at rest

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!