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 using symbolic algebra.

(c) What is the numerical value for M/m?

(d) What is the value of teh tangential accerlatio of a point on the rim of the horizontal disk?

Explanation / Answer

mass of the disk and cylinder is M

radius of the disk and cylinder is R

radius of the pulley, r=10*R

and

the mass which falling , m=M/9

a)

moment of inertia,

I_total=I_disk + I_cylinder

=1/2*M*R^2 + M*R^2

=(3/2)*M*R^2

b)

for mass, m

T=m*g-m*a

T=(g-a)*m

for the disk,

T*r=I*alpa

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

but,

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

(g-a)*m*10=(3/2)*M*(a/10)

(g-a)*m=(3/200)*M*a

==>

M/m=(3/200*(a/(g-a))

==>

m/M=(200/3)*(g/a-1)

g/a-1=(3m/200*M)

g/a=1+3m/(200*M)

a=(g/(1+3m/(200M))

c)

from the above part (b)

M/m=(3/200*(a/(g-a))

d)

a_tan=R*alpa

=R*(a/r)

=R*(a/10*R)

=a/10

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