Pivoting Cylinder?

Question

A solid cylinder (M=2.43 kg, R=0.131 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 0.570 kg mass, i.e., F = 5.592 N. Calculate the angular acceleration of the cylinder.

If instead of the force F an actual mass m = 0.570 kg is hung from the string, find the angular acceleration of the cylinder.

How far does m travel downward between 0.430 s and 0.630 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 0.381 m in a time of 0.510 s. Find Icm (I sub cm) of the new cylinder.

Answer 1

F=ma
T=I A
T – Torque
I – moment of inertia (for cylinder in this case I=.5mR^2)
A – Angular acceleration (A= a/R where (a) is tangential acceleration and ® is the radius)

T=R x F (also torque equal force times the perpendicular component of the arm or radius)
I=.05 m R^2
A=T/I= 2 R F/ (m R^2)
A=2 F/(mR)
A=2 x 5.592 /(2.43 x 0.131 )=
A=35 rad/sec^2 this is for a constant force of 5.592

Part (b)
Consider the system .

F(total)=W(weigh tof the mass) - F(due to cylinder's inertia)
or
m1a=m1g - IA/R
A=a/R (tangential acceleration of th ecylinder is the same as the mass of the weight)
m1a=m1g - .5 m2 R^2 (a/R)/R

Now we have (after cleaning up the mess);
a=(m1 g )/ (m1 + .5 m2 ) this is acceleration of the system
a=(0.570 x 9.81)/(0.570 + 0.5 x 2.43 )
a = 3.1 m/sec^2

Angular acceleration of the cylinder A
A=a/R=(m1 g )/ [R(m1 + .5 m2 ) ]
A=23.9 rad/sec^2

Part (C)
h=.5at^2 (Initial velocity =0)
h2-h1=.5a(t2^2 - t1^2)=
h2-h1=0.5 x 3.1( (0.630 )^2 - (0.430 )^2)
h2-h1=0.3286 m

Part (D)
The cylinder is changed to one with the same mass and radius, but a different moment of inertia.
Hmm.. Is that possible?
since I=(mR^2)/2

Answer 2

Torque, T = RF

But T is also = Iα, therefore

α = RF/I, where I is the moments of inertia for a solid cylinder spinning about the z-axis, so I = ½MR²

α = 2RF/MR² = 2F/MR = 2(5.592)/(2.43)(0.131)

= 35.13 radians per sec²

If we consider an actual mass, m hanging from the string, instead of the force F, then we must consider the moments of inertia of mass m as well:

α = RF/I = Rmg/(Ic + Im)

where Ic is the moments of inertia of the solid cylinder, and
Im is the moments of inertia of mass m

Ic = ½MR², same as before

Im = mR², here we assume that mass m is like a point mass and is placed at a distance R from the center of the cylinder before it is dropped.

α = mgR/(Ic + Im) = mgR / (½MR²+mR²) = 2mg/R(M+2m)

= 2(0.57)(9.81) / (0.131)(2.43 + 2(0.57))

= 23.91 radians per sec²

Between t = 0.430 sec and 0.630 sec after the motion begins, the mass m travels down:

θ(t) = ½αt²

so θ(t=0.63) = (0.5)(23.91)(0.63)² = 4.746 radians

and θ(t=0.43) = (0.5)(23.91)(0.43)² = 2.21 radians

So at t = 0.43 sec, mass m travels down

d = θR = (2.21)(0.131) = 0.29 m

and at t = 0.63 sec, mass m travels down

d = θR = (4.746)(0.131) = 0.62 m

So between t = 0.43 and 0.63 sec, mass m travelled

Δd = 0.62 - 0.29 = 0.33 m

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To find Icm, the new moments of inertia of the cylinder, you just work backwards.

Use this θ(t) = ½αt² to find α, and then use

α = mgR/(Icm + Im) to find Icm