During the Moment of Inertia experiment, a group of students decided to niount c

Question

During the Moment of Inertia experiment, a group of students decided to niount cylindrical shell on the top of the horizontal disk. The cylindrical shell and disk have the same mass, M, and radius, R. The friction was removed from the system. The radius of the pulley, 7, was exactly 1/10 of the radius o the lincar inertia of the disk and cylindrical shell together. (b) Find the ratio of the mass of the disk to the mass of the falling mass, M/m using smbolic algebra f the disk and acceleration of the falling mass was measured to be 16.25 mm/s?. (a) Calculate the moment of o What is the numerical value for M/m () What is tangential acccteration of a point on the rim of the horizontal disk? Recall that: g980m/sIorizostal dskMR IphereMR, lytinrMand ylindrical shellMR olating jo t aling m

Explanation / Answer

mass of the disk and shell is M

radius of the disk and shell is R

radius of the pulley, r=R/10

acceleration, a=16.25*10^-3 m/sec^2

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

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

(g-a)*m=(300/2)*M*a

==>

M/m=(2/300*(g-a)/a)

c)

M/m=(2/300*(g-a)/a)

M/m=(2/300*(9.8-16.25*10^-3)/(16.25*10^-3)

M/m=4.014

d)

a_tan=R*alpa

=R*(a/r)

=R*(a*10/R)

=10*a

=10*16.25*10^-3

=0.1625 m/sec^2

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