A large cylinder of radius r 2 = 0.437m has a hole through the center of radius

Question

A large cylinder of radius r2= 0.437m has a hole through the center of radius r1= 0.185m. The solid region of the bored out cylinder has a uniform mass density with a total mass m = 19.6kg and it has a length of L = 1.85m. The moment of inertia when the cylinder spins about the center of mass at L/2 is shown in the above diagram is derived above. (a) What is the torque on the cylinder when we apply an angular acceleration of ? = 3.32rad/s2? (b) If the cylinder is subject to the angular acceleration for t = 1.32s, then what is the final angular velocity?

Answers for (a) kg m^2/s^2 (b) rad/s^2

Explanation / Answer

I = 19.6 / 4 * ( 0.437^2 + 0.185 ^2 ) + (19.6 / 12) * 1.85^2 = 6.69 kg m^2............ torque T = I angular accelaration = 6.69 * 3.32 = 22.22 Nm........... b ) angular velocity = angular accelaration * time = 4.38 rad / s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.