a)M, a solid cylinder (M=2.35 kg, R=0.117 m) pivots on a thin, fixed, frictionle

Question

a)M, a solid cylinder (M=2.35 kg, R=0.117 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 0.630 kg mass, i.e., F = 6.180 N. Calculate the angular acceleration of the cylinder.

b)If instead of the force F an actual mass m = 0.630 kg is hung from the string, find the angular acceleration of the cylinder.

c) How far does m travel downward between 0.570 s and 0.770 s after the motion begins?

d) 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 0.282 m in a time of 0.490 s. Find Icm of the new cylinder.

Explanation / Answer

In rotational mechanics, we know that Torque= I*a' ( where a'= angular accleration of cylinder).............(1)

now Moment of inertia of solid cylinder, I= (1/2)MR2= (1/2)(2.35)(0.117)2= 0.0160 kgm2.................(2)

Force exerted on cylinder, F= 6.180 N

Torque on cylinder, T= F*R ( as angle b/w force and radius is 90 degrees)

thus, T= (6.18)*(0.117)= 0.723 Nm

Using equation 1 and 2

T= Ia'.............................(5)

angular acceleration, a'= T/I= 0.723/0.016= 45.187 rad/s2 (ANS)

b). If mass of same weight is hanged on string then also same force will exert on the cylinder and thus angular acceleration will be same.

c). linear accleration, a= (a')*R

thus from given and calculated values, a= (45.187)(0.117)= 5.286 m/s2

Initially it started from rest, Hence u= 0 m/s

Let in time t1= 0.570 s, Distance travelled by it is S1 then by using equations of motion

S1= ut1+(1/2)at12

S1= 0+(1/2)(5.28)(0.570)2= 0.858 m.................(3)

Similarly distance travlled in time t2= 0.770 s be S2 then

S2= ut2+(1/2)at22

S2= 0+(1/2)(5.28)(0.770)2= 1.565 m.........................(4)

Thus Distance travlled b/w 0.570s and 0.770 s is S= S2-S1= (1.565-0.858)m= 0.707 m

d). Given S= 0.282 m in time t= 0.490 s. Again initial velocity , u= 0m/s ( startng from rest)

Let new accelration be an

thus, by using equations of motion

S= ut+(1/2)ant2

   0.282=0+(1/2)an(0.490)2

an= (2*0.282)/0.240= 2.35 m/s2

Also let new angular acceleration be "a'n" due to the same mass or same force or same torque.

thus, Linear accleration,an= a'n* R

or using equation 5,

an= (T/In)*R

New moment of inertia , In= T*R/an= (0.723*0.117)/2.35= 0.0359 kgm2 (ANS)

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