M, a solid cylinder (M = 2.03 kg, R = 0.111 m) pivots on a thin, fixed, friction

Question

M, a solid cylinder (M = 2.03 kg, R = 0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.590 kg mass, i.e., F = 5.788 N. Calculate the angular acceleration of the cylinder. If instead of the force F an actual mass m = 0.590 kg is hung from the string, find the angular acceleration of the cylinder. How far does m travel downward between 0.690 s and 0.890 s after the motion begins? The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.318 m in a time of 0.470 s. Find I_cm of the new cylinder.

Explanation / Answer

mass M=2.03 kg,

radius R=0.111 m

force =weight =0.59*9.8=5.788 N

a)

let,

angular acceleration is alpa

torque =I*alpa

R*F*sin(theta)=I*alpa

R*F*sin(90)=(1/2)*(M*R^2)*alpa

R*F=(1/2)*(M*R^2)*alpa

alpa=2*F/M*R

alpa=2*5.788/(2.03*0.111)

alpa=51.373 rad/sec^2

b)

now, mass m=0.59 kg

tension T=m*g-m*a

and

a=R*alpa

now,

Torque =I*alpa

R*T=1/2*M*R^2*alpa

===>

alpa=2*T/M*R

alpa=2*(m*g-m*alpa*R)/M*R

===>

alpa=2*m*g/(R(M+2m)

alpa=(2*0.59*9.8)/(0.111(2.03+2*0.59)

======>

alpa=32.455 rad/sec^2

c)

let,

t1=0.69 sec and t2=0.89 sec

distance travelled,

s=1/2*a*(t2^2-t1^2)

s=1/2*(R*alpa)*(t2^2-t1^2)

s=1/2*0.111*32.455*(0.89^2-0.69^2)

s=0.57 m

d)

distance s=0.318 m

time t=0.47 sec

now again,

torque,T*R=I*alpa

T*R=I*(a/R)

I=T*R^2/a

here,

s=1/2*a*t^2

0.318=1/2*a*0.47^2

===> a=2.88 m/sec^2

T=m*g-m*a

T=(0.59*9.8)-(0.59*2.88) =4.083 N

now,

I=T*R^2/a

I=4.083*0.111^2/2.88

I=0.0175 kg.m^2

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