How Is Rotational Kinetic Energy Calculated for Two Balls on a Rod?

[dsptl]

A 300 g ball and a 500 g ball are connected by a 60-cm-long massless rigid rod. The structure rotates about the center of the rod with angular velocity 5 rad/s. What is the rotational kinetic energy?

Attempt:
the rotational kinetic energy is given by
KEr = (1 / 2) I ω^2
where I is the moment of inertia if the ball is considered as a solid sphere then we get that
I = (2 / 5) M R^2 for a solid sphere


but i have a question what should i use for M and R?

 

[christensen]

you don't have a solid sphere here.

0------X------0

where the "0"s are the balls and the "X" is the axis of rotation.

 

[dsptl]

you don't have a solid sphere here.

0------X------0

where the "0"s are the balls and the "X" is the axis of rotation.

so wht should i do?

 

[christensen]

you want to calculate the moment of inertia of the system here and continue as you were planning on with E = 1/2 * I * w^2

 

[dsptl]

you want to calculate the moment of inertia of the system here and continue as you were planning on with E = 1/2 * I * w^2

for I its just .5MR^2

and what would be my M and R since there r two mass given?

 

[christensen]

Moment of Inertia is the sum of all of (mass times the sqaure of the distance to the axis of rotation)

In other words I =$$\sum$$ m_ir_i²

 

[dsptl]

so I = (.5 + .3 ) (.60)^2

T = Ia

and solve for a, right?

 

[christensen]

if the rotational axis is in the center of the rod, and masses are on the ends, and the rod is 60cm long, is the distance between the axis of rotation and the mass 60cms?

 

[dsptl]

so I = (.5 + .3 ) (.30)^2

T = Ia

and solve for a, right?

 

[christensen]

im not sure what you mean by "T = Ia"

 

[dsptl]

oh sorry its suppose to E = .5Iw^2

and solve for E