Falling Block Pulling Rolling Cylinder

[Bluestribute]

Homework Statement

A uniform, solid cylinder of mass M = 4.22 kg and radius R = 0.36 m with I=1/2(M*R^2) is attached at its axle to a string. The string is wrapped around a small frictionless pulley (I=0) and is attached to a hanging block of mass 1.69 kg. You release the objects from rest. Assume the cylinder rolls without slipping. What is the acceleration of the block

Homework Equations

I=1/2(M*R^2)
a(x)=a(y)

The Attempt at a Solution

block
Tension=M(b)*(g-a)

cylinder
∑X: Tension-ƒs=M(c)*a
∑Torque (C.O.M in center of cylinder): ƒsR+M(c)g2R=I$\alpha$

I'm assuming, on the cylinder, normal and weight offset, so N=M(c)g, and because the normal makes contact at the bottom of the cylinder, the distance is the diameter (or 2R in the torque calculation).

ƒs=Tension-M(c)*a

Back to Torque, subbing for tension and static friction, inertia and angular acceleration in terms of linear acceleration (and canceling out on the I$\alpha$ side):

(M(b)*(g-a))R-(M(c)*a*R)+M(c)g2R=(M(c)*a)/2
M(b)gR + M(c)g2R = a((M(c)/2) + M(b)R + M(c)R)

So, substituting in the given numbers, I get:

((1.69*9.81*0.36) + (4.22*9.81*0.72)) / ((0.5*4.22) + (1.69*0.36) + (4.22*0.36)) = a

Basically, I solve for tension using the block. Then I solve for static friction using the sum of forces in the X direction of the cylinder. Then I solve for linear acceleration using torques of the cylinder, which is just the normal (with a radius of 2R) and static friction. Answers I've gotten somehow:

1 Incorrect. (Try 1) 0.36 m/s/s
2 Incorrect. (Try 2) 36.96 m/s/s
3 Incorrect. (Try 3) 2.62 m/s/s
4 Incorrect. (Try 4) 1.41 m/s/s
5 Incorrect. (Try 5) 2.13 m/s/s
6 Incorrect. (Try 6) 2.67 m/s/s
7 Incorrect. (Try 7) 8.44 m/s/s

 

[Simon Bridge]

7 tries with wildly different values - you have made a mistake in your algebra or reasoning. Are you sure you understand the physics involved?

Note: in

M(b)gR + M(c)g2R = a((M(c)/2) + M(b)R + M(c)R)​

... the dimensions don't match. Typo?

Recheck and go step by step - likely place to mess up is in the treatment of "rolling without slipping".

Try simplifying your notation to make your maths clearer - say make m the falling mass and M the cylinder mass: avoids subscripts.

I think you have (at least) misplaced a minus sign - check that your vector directions are consistent.
I'd choose the +ve direction as the direction of the linear acceleration you expect.

 

[ehild]

The Attempt at a Solution

block
Tension=M(b)*(g-a)

cylinder
∑X: Tension-ƒs=M(c)*a
∑Torque (C.O.M in center of cylinder): ƒsR+M(c)g2R=I$\alpha$

I'm assuming, on the cylinder, normal and weight offset, so N=M(c)g, and because the normal makes contact at the bottom of the cylinder, the distance is the diameter (or 2R in the torque calculation).

The equation in red is wrong.
If you consider the torque with respect to the C.O.M you do not have the second term.
If you consider the torque with respect to the instantaneous axis, the torque of fs is zero, and the torque of gravity is not M(c)2gR and the moment of inertia is not 0.5 M(c) R².

ehild

 

[Bluestribute]

Really? If the C.O.M is at the center, and the second term is the normal, wouldn't it still count for torque since the contact isn't at the C.O.M., but rather the ground? Or does that not matter because it goes through the center anyways? Lemme try setting static friction equal to torque (?), and if that doesn't work and I still need help, I rewrite everything so it's easier to follow.

 

[Bluestribute]

Ok, now I got the same answer twice in a row so where am I going wrong now?

M$_{}c$ = Mass Cylinder = 4.22 kg
M$_{}b$ = Mass Block = 1.69 kg
R = 0.36 m
I = 0.5*M*R$^{}2$

∑$\tau$ = ƒs*R = I*(a/R)
ƒsR = Fτ - M$_{}c$*a

Fτ*R - M$_{}c$*a*R = I*(a/R)

Now, using the block to get the tension:
Fτ = M$_{}b$*(g+a)

Substituting that into the torque equation:
M$_{}b$*(g+a)*R - M$_{}c$*a*R = I*(a/R)

Now getting acceleration onto one side (and factored out):
M$_{}b$*g*R = a((I/R) + M$_{}c$*R - M$_{}b$*R)

What am I doing wrong?

 

[ehild]

Really? If the C.O.M is at the center, and the second term is the normal, wouldn't it still count for torque since the contact isn't at the C.O.M., but rather the ground? Or does that not matter because it goes through the center anyways? Lemme try setting static friction equal to torque (?), and if that doesn't work and I still need help, I rewrite everything so it's easier to follow.

Really. Your torque equation is wrong. Decide what is the axis with respect you determine the torque of all forces. You can not calculate the torque of one force with respect to one axis, the torque of the other force with respect to an other axis and then add them.

ehild