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Mass falling and pulling a cylinder without sliding

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  • #1
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Homework Statement


A cylinder of mass M and radius r is at distance l0 from the edge. mass m is hanging from a rope. no friction at the weightless pulley. the cylinder starts rotating without sliding.
What's the acceleration of the COM of the cylinder.
What's the velocity of the COM of the cylinder when it reaches the edge.
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Homework Equations


Torque and acceleration of a rigid body: ##M=I\alpha##
Moment of inertia of a solid cylinder: ##I_c=\frac{mr^2}{2}##
Shteiner's theorem: ##I_b=I_a+md^2##

The Attempt at a Solution


Relation between angular and linear acceleration of COM:
$$2r\alpha=\dot{x_A}\;\rightarrow\; \alpha=\frac{\dot{x_A}}{2r}$$
Tension in rope is T:
$$\left\{ \begin{array}{l} mg-T=m\dot{x_A} \\ 2rT=(I_c+Mr^2)\frac{\dot{x_A}}{2r} \end{array}\right.$$
$$\rightarrow \dot {x_A}=\frac{4mgr^2}{I_c+(M+4m)r^2}$$
The acceleration of the COM: ##\dot{x_c}=\frac{1}{2}\dot{x_A}##
The velocity at the edge:
$$v^2=2\dot{x_c}l_0\;\rightarrow\; v^2=\dot{x_A}l_0$$
 

Answers and Replies

  • #2
ehild
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Homework Statement


A cylinder of mass M and radius r is at distance l0 from the edge. mass m is hanging from a rope. no friction at the weightless pulley. the cylinder starts rotating without sliding.
What's the acceleration of the COM of the cylinder.
What's the velocity of the COM of the cylinder when it reaches the edge.
View attachment 100056

Homework Equations


Torque and acceleration of a rigid body: ##M=I\alpha##
Moment of inertia of a solid cylinder: ##I_c=\frac{mr^2}{2}##
Shteiner's theorem: ##I_b=I_a+md^2##

The Attempt at a Solution


Relation between angular and linear acceleration of COM:
$$2r\alpha=\dot{x_A}\;\rightarrow\; \alpha=\frac{\dot{x_A}}{2r}$$
Think it over.
 
  • #3
haruspex
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Presumably everywhere you wrote ##\dot x_A## you mean ##\ddot x_A##, xA being the displacement of m.
ehild seems to take exception to your equation relating that to angular acceleration, but it looks right to me.
I agree with your answers, but you should simplify them using the formula you quote for Ic.
 
  • #4
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The angular acceleration should be r instead of 2r
 
  • #5
ehild
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I see, the equations are written for the acceleration of point A, with respect to the point of contact.
The problem can be solved by conservation of energy, too, taking into account that the vertical displacement (and velocity) of m is twice the horizontal displacement/velocity of the CM.
 
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  • #6
haruspex
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The angular acceleration should be r instead of 2r
r is a radius, not an acceleration. What do you mean?
 
  • #7
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$$2r\alpha=\ddot x_A \;\rightarrow\; \alpha=\frac{\ddot{x}_A}{2r}$$
$$\left\{ \begin{array}{l} mg-T=m\ddot x_A \\ 2rT=(I_c+Mr^2)\frac{\ddot x_A}{2r} \end{array}\right.$$
$$\rightarrow \ddot x_A=\frac{4mgr^2}{I_c+(M+4m)r^2}$$
$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+2m}$$
$$v^2=2\ddot x_c l_0\;\rightarrow\; v^2=\ddot x_A l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+2m}}$$
The problem can be solved by conservation of energy
$$mgl_0=\frac{1}{2}I_B\omega^2=\frac{1}{2}\left( I_c+Mr^2 \right)\omega^2=\frac{\omega^2}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]$$
$$\rightarrow \omega^2=\frac{4mgl_0}{3Mr^2}$$
$$v=\omega r=...=2\sqrt{\frac{mgl_0}{3M}}$$
They differ
 
  • #8
ehild
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They differ
You forgot the KE of the hanging mass. Lo is the distance traveled by the CM. There are mistakes in the other derivation, too.
 
  • #9
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$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}(I_C+Mr^2)\omega^2$$
$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$
 
  • #10
ehild
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$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+2m}$$
When distributing factor 2 in the denominator, you divided 4m by 2 instead of multiplying.
$$v^2=2\ddot x_c l_0\;\rightarrow\; v^2=\ddot x_A l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+2m}}$$
It would be otherwise correct.
 
  • #11
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$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+8m}$$
$$v^2=2\ddot x_c l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+8m}}$$
But still differs from the energy method in post #9 which gave:
$$v=2\sqrt{\frac{mgl_0}{M}}$$
 
  • #12
ehild
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$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
I do not follow your notations. The KE of the cylinder is the sum of those of translational and rotational motion. If the CM travels with speed V, ω=V/r. KE(cylinder)= 3/4MV2.
The falling block moves with speed of 2V, and the change of its height is 2l0.
 
  • #13
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Ive assumed that linear distance moved by the hanging mass and the cylinder will be the same.
The driving force (F) for the system is = m * g (Newtons)
Assuming its a solid cylinder rolling without slipping, then in this problem, its effective mass = 1.5 * M
So the linear acceleration (a) is from : F / ( ( 1.5 * M ) + m )
 
  • #14
haruspex
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Ive assumed that linear distance moved by the hanging mass and the cylinder will be the same.
Not a good start.
 
  • #15
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In:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
v is the speed of the falling mass, it's kinetic energy contribution. on the right side, the ##\frac{1}{2}I_B\omega^2## is the combined translational+rotational energy of the cylinder.
Later, in:
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$
The v is for the center of the cylinder. indeed not a wise notation, but i assumed it will be understood.
 
  • #16
ehild
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In:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
v is the speed of the falling mass, it's kinetic energy contribution. on the right side, the ##\frac{1}{2}I_B\omega^2## is the combined translational+rotational energy of the cylinder.
Later, in:
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$
The v is for the center of the cylinder. indeed not a wise notation, but i assumed it will be understood.
But your result is wrong. You included the KE of the falling block in the first equation, but ignored later.
 
  • #17
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You included the KE of the falling block in the first equation, but ignored later.
I don't have and don't know how to include the KE of the falling block later. i made:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
The connection between v, the velocity of the falling block at the edge of the table, and ω, the angular velocity at the edge is ##v=2\omega r##, and i am left only with ω.
Now i translate ω that i found into vc by ##v_c=\omega r##. where and how to take into consideration KE?
 
  • #18
ehild
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I don't have and don't know how to include the KE of the falling block later. i made:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
The connection between v, the velocity of the falling block at the edge of the table, and ω, the angular velocity at the edge is ##v=2\omega r##, and i am left only with ω.
When you collect the terms with ω2, there should be a term with m. But it is missing from your formula ##\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}}##
Now i translate ω that i found into vc by ##v_c=\omega r##.
That is correct.
 
  • #19
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When you collect the terms with ω2, there should be a term with m
$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$\left( \frac{3}{4}Mr^2-2mr^2 \right)\omega^2=-mgl_o\;\rightarrow\; \omega=\frac{2}{r}\sqrt{\frac{-mgl_0}{3M-8m}}$$
Not good, negative under the square root.
 
  • #20
haruspex
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$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
That says the gain in energy of the descending mass equals the gain in energy of the rotating cylinder. I'm sure you did not mean that.
 
  • #21
ehild
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$$
-mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$
Not good, negative under the square root.
Your energy balance equation is wrong. The KE increases, the PE decreases. The total energy remains constant. ΔKE + ΔPE =0
What is ΔPE? What is the displacement of the hanging mass if the CM of the cylinder is displaced by lo?
 
  • #22
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$$W=-\Delta E_P=-(E_{pf}-E_{pi})=\Delta E_K$$
$$-(-mgl_0)=\frac{1}{2}mv^2+\frac{1}{2}I_B\omega^2,\; v=2\omega r$$
$$mgl_0=\left( 2m+\frac{3}{4}M \right)r^2\omega^2\;\rightarrow\; \omega=\frac{2}{r}\sqrt{\frac{mgl_0}{3M+8m}}$$
$$v_c=\omega r\;\rightarrow\; v_c=2\sqrt{\frac{mgl_0}{3M+8m}}$$
Like in the kinematic approach in post #11
 
  • #23
ehild
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$$W=-\Delta E_P=-(E_{pf}-E_{pi})=\Delta E_K$$
$$-(-mgl_0)=\frac{1}{2}mv^2+\frac{1}{2}I_B\omega^2,\; v=2\omega r$$
$$mgl_0=\left( 2m+\frac{3}{4}M \right)r^2\omega^2\;\rightarrow\; \omega=\frac{2}{r}\sqrt{\frac{mgl_0}{3M+8m}}$$
$$v_c=\omega r\;\rightarrow\; v_c=2\sqrt{\frac{mgl_0}{3M+8m}}$$
Like in the kinematic approach in post #11
Not quite. Compare them. There is still problem with the PE.
 
  • #24
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Cylinder linear speed is half of the falling mass, ok :
The acceleration of the falling mass = ( m * g ) / ( ( m * 0.375 ) + M )
The acceleration of the cylinder is half that value.
 
  • #25
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$$-(-2mgl_0)=\frac{1}{2}mv^2+\frac{1}{2}I_B\omega^2,\; v=2\omega r$$
$$2mgl_0=\left( 2m+\frac{3}{4}M \right)r^2\omega^2\;\rightarrow\; \omega=\frac{2}{r}\sqrt{\frac{2mgl_0}{3M+8m}}$$
$$v_c=\omega r\;\rightarrow\; v_c=2\sqrt{\frac{2mgl_0}{3M+8m}}$$
 

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