During the Moment of Inertia experiment, group of students decided to mount soli

Question

During the Moment of Inertia experiment, group of students decided to mount solid cylinder on the top of the horizontal disk. The cylinder and disk have the same mass, M, and radius, R. The friction was removed from the system. The falling mass, Ill, was 1/9 of the mass of the disk and the radius of the solid cylinder is 10 times greater than radius of the pulley. (a) Calculate the moment of inertia of the disk and solid cylinder together. (b) Find the linear acceleration of the falling mass. (c) What is the value of tangential acceleration of a point on the rim of the horizontal disk? Please give a detailed description.

Explanation / Answer

moment of inertia = Idisk + Icylinder = (1/2)*M*R^2 + M*R^2 = (3/2)*M*R^2

(b)

for m

mg - T = m*a

T = mg - ma

for disk

T *r = I*alpha

angular acceleration alpha = a/r

(m*g - ma)*r = (3/2)*M*R^2*a/r

given r = 10*R

m = M/9

(M/9*g - M/9*a)*10*R = (3/2)*M*R^2*a/10R

Mg - Ma = 9*(3/20)*M*a

g - a = (27/20)*a

g = 47/20 a

a = g*(20/47) = 9.8*(20/47) = 4.17 m/s^2 <<<<------answer

=========================

atan = R*alpha = R*a/r = R*a/(10R) = a/10 = 0.417 m/s^2 <<<<-answer

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

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