A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an

Question

A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m_1 = 5.10 kg is attached to one end and a second mass m_2 = 3.40 kg is attached to the other end of the rod. Treat the two masses as point particles. What is the moment of inertia of the system? What is the axis of rotation for the rod? Do the two end masses contribute to the moment of inertia of the rotating system? kg middot m^2 If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have? Consider the basic definition of rotational kinetic energy. J Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined? If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Explanation / Answer

solution:

A) Isystem =Irod +Iblocks

Irod =ML2 /12 ;about its centre

Irod =(2.30)(22) /12 = .7666 Kg.m2

now for blocks

Iblock1 =MR2 =(5.10Kg)(12) ;here R= 1m about its centre of rotation

Iblock1 =5.10

similarly for block 2

Iblock2 =3.40

so I system =.7666+5.10+3.40

=9.26Kg.m2 ans A

B)

Kinetic energy =.5*Isystem*w2

here w =2.70 rad/s

so K.E =.5*9.266*2.702 =33.77 J ans B

C)

now rod mass is negligible

so Isystem =Iblocks =5.10+3.40 =8.50 Kg.m2 ans

D)

K.E =.5*(8.50)(2.702) =30.98 J ans

  

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