Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A uniform rod of mass 2.20 kg and length 2.00 m is capable of rotating about an

ID: 1291044 • Letter: A

Question

A uniform rod of mass 2.20 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 4.80 kg is attached to one end and a second mass m2 = 2.70 kg is attached to the other end of the rod. Treat the two masses as point particles. (a) What is the moment of inertia of the system? (b) If the rod rotates with an angular speed of 2.40 rad/s, how much kinetic energy does the system have? (c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.40 rad/s?

A uniform rod of mass 2.20 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 4.80 kg is attached to one end and a second mass m2 = 2.70 kg is attached to the other end of the rod. Treat the two masses as point particles. (a) What is the moment of inertia of the system? (b) If the rod rotates with an angular speed of 2.40 rad/s, how much kinetic energy does the system have? (c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.40 rad/s?

Explanation / Answer

a) Irod = m L2 /12 = 2.2 * 22 / 12 = 0.73 kg.m2

Ip1= m r2= 4.8 * 12 = 4.8 kg.m2

Ip2 = m r2 = 2.7 * 12 = 2.7 kg.m2

Total Moment of inertia

= 0.73 + 4.8 + 2.7 = 8.23 kg.m2

--------------------------------------------------

b) KE = 1/2 * I * w2

= 1/2 * 8.23 * 2.402= 23.7 J

--------------------------------------------------

c)I = 0 + 4.8 + 2.7 = 7.5 kg.m2

-------------------------------------------------

d) KE = 1/2 * I * w2 = 1/2 * 7.5 * 2.42

= 21.6 J

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote