The rigid object shown is rotated about an axis perpendicular to the paper and through point P.The total kinetic energy of the object as it rotates is equal to 1.4 J.If M = 1.3 kg and L = 0.50 m,what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles.
A)1.3 rad/s
B)1.5 rad/s
C)1.7 rad/s
D)1.2 rad/s
E)2.1 rad/s

Question

The rigid object shown is rotated about an axis perpendicular to the paper and through point P.The total kinetic energy of the object as it rotates is equal to 1.4 J.If M = 1.3 kg and L = 0.50 m,what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles. 
A)1.3 rad/s 
B)1.5 rad/s 
C)1.7 rad/s 
D)1.2 rad/s 
E)2.1 rad/s

Quizplus · Accepted Answer

The kinetic energy of a rotating object can be expressed as: K = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.

To find the moment of inertia of the object, we can use the parallel axis theorem. The moment of inertia of the object about an axis passing through its center of mass (which is also the axis perpendicular to the paper and passing through point P) is:

I_cm = I_rod + 2(I_disk + MR^2)

where I_rod = (1/3)ML^2 is the moment of inertia of the rod about its center of mass, I_disk = (1/2)MR^2 is the moment of inertia of each disk about its own center of mass, and R is the radius of each disk.

Using the given values, we have:

I_cm = (1/3)(1.3 kg)(0.5 m)^2 + 2[(1/2)(0.25 kg)(0.1 m)^2 + (0.25 kg)(0.25 m)^2]
     = 0.3825 kg m^2

Now, we can use the given kinetic energy to solve for the angular velocity:

K = (1/2)Iω^2

1.4 J = (1/2)(0.3825 kg m^2)ω^2

ω^2 = 7.3168

ω = 1.707 rad/s (rounded to three significant figures)

Therefore, the answer is (C) 1.7 rad/s.

The Rigid Object Shown Is Rotated About an Axis Perpendicular