Calculating K.E: 0.0136 J Solution

[sfx81]

Hi, I am trying to calculate Kinetic energy of system where a mass m(0.20 kilogram) attached at one ond of turntable rotating at 1 revolution per second. The radius of turntable is 6cm.
Here I solve it, please correct

Kinetic Energy = (1/2)mv²

calculating velocity at 1 rpm, and radius = 6cm
so for circumfrance of circle = 2*pi*r
v = 2 * 3.14 * 6cm = 37.69 cm = 0.3769 meters/second

K.E = 0.5 * 0.20 * 0.37² meters
= 0.0136 Joules <=== Is this correct ?

Any help will be appreciated.
Thanks

 

[Kalvarin]

You need to work out it's rotational kinetic energy.

I = m*(r^2)

I is the moment of inertia of a point mass rotating about a fixed point. Mass of turntable isn't mentioned so assume it is negligible.

E = 0.5*I*(w^2)

E is the energy of a rotating system, where w is angular frequency which is found by:

w = 2*($$\pi$$)/T

 

[sfx81]

Hi . So If I get it right
mass = 0.02Kg, radius = 6cm = 0.06 meter, frequency = 1 revolution per second
KE_rotation = I * w²
where w = angular velocity = 2 * pi * f , where f = frequency in revolutions per second.
I = m * r²
K.E _rotation = m * r² * ( 2 * pi * 1)²
= 0.02 * 0.06² * ^(2*pi*1)
= 2.841 * 10^-3 Joules

If this is correct, then how do we equate in the mass of turn table (e.g 0.5 Kg) into equation ? Cause rotational kinetic energy deals with point mass around rotation axis, where as a turn table alone has uniform mass distribution along its surface ?

Any help will be appreciated.
Thanks