*** I figured out all the answers for this problem except for the last 3 (h,i,J)

Question

*** I figured out all the answers for this problem except for the last 3 (h,i,J), PLEASE HELP***

An oversized yo-yo is made from two identical solid disks each of mass

M = 2.40 kg and radius R = 10.1 cm. The two disks are joined by a solid cylinder of radius r = 4.00 cm

and mass m = 1.00 kg as in the figure below. Take the center of the cylinder as the axis of the system, with positive torques directed to the left along this axis. All torques and angular variables are to be calculated around this axis. Light string is wrapped around the cylinder, and the system is then allowed to drop from rest.

(a) What is the moment of inertia of the system? Give a symbolic answer. (Use any variable or symbol stated above as necessary. Do not substitute numerical values; use variables only.)

I = (MR2)+ (mr2/2)

(b) What torque does gravity exert on the system with respect to the given axis?

0 N · m

(c) Take downward as the negative coordinate direction. As depicted in the figure above, is the torque exerted by the tension positive or negative?

positive

Is the angular acceleration positive or negative?

positive

What about the translational acceleration?

negative

(d) Write an equation for the angular acceleration (alpha) in terms of the translational acceleration a and radius r. (Watch the sign!)

a= -a/r

(g) Eliminate from the rotational second law with the expression found in part (d) and find a symbolic expression for the acceleration a in terms of m, M, g, r and R.

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

(h) What is the numeric value for the system's acceleration?

? m/s2

(i) What is the tension in the string?

? N

(j) How long does it take the system to drop 1.20 m from rest?

?s

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

Explanation / Answer

given,

M=2.4 kg

m=1kg

R=10.1 cm

r=4cm

h)

acceleration of the system is,

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

=(2*2.4+1)9.8/(2*2.4+2.4(10.1/4)^2+3*1/2)

a=2.63 m/sec^2

i)

torque =I*alpa

T*r=[(M*R^2)+(m*r^2)/2]*(a/r)

T=[(M*(R/r)^2)+m/2]*a

=[(2.4*(10.1/4)^2)+1/2]*2.63

=17.25 N

j)

distance s=1.2 m

no of roatation n=1.2/(2*3.14*0.04) =4.78 rotations

now,

angular dispalacement theta=2pi*4.78

=30 rad

and

theta=1/2*alpa*t^2

30=(1/2)*(2.63/0.04)*t^2

====>

t=0.955 sec

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