Mass falling and pulling a cylinder without sliding

• Karol
In summary: KE of the block in the second - the one that you use to find omega. It should be 3/2 mv^2 there, not 1/2 mv^2.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 * MSo the linear acceleration (a) is from : F / ( ( 1.5 * M ) + m )Sorry, but this is not a correct approach. The cylinder is accelerating in a tangential direction, hence the acceleration of the CM is not equal to the acceleration of the hanging mass. Also, the effective mass
Karol

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.

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:
$$v^2=2\dot{x_c}l_0\;\rightarrow\; v^2=\dot{x_A}l_0$$

Karol said:

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.

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.

The angular acceleration should be r instead of 2r

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.

Last edited:
kinemath said:
The angular acceleration should be r instead of 2r
r is a radius, not an acceleration. What do you mean?

$$2r\alpha=\ddot x_A \;\rightarrow\; \alpha=\frac{\ddot{x}_A}{2r}$$
$$\left\{ \begin{array}{l} mg-T=m\ddot x_A \\ 2rT=(I_c+Mr^2)\frac{\ddot x_A}{2r} \end{array}\right.$$
$$\rightarrow \ddot x_A=\frac{4mgr^2}{I_c+(M+4m)r^2}$$
$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+2m}$$
$$v^2=2\ddot x_c l_0\;\rightarrow\; v^2=\ddot x_A l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+2m}}$$
ehild said:
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

Karol said:
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.

$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}(I_C+Mr^2)\omega^2$$
$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$\rightarrow\omega=\frac{2}{r}\sqrt{\frac{mgl_0}{M}},\;\; v=\omega r\;\rightarrow\; v=2\sqrt{\frac{mgl_0}{M}}$$

Karol said:
$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+2m}$$
When distributing factor 2 in the denominator, you divided 4m by 2 instead of multiplying.
Karol said:
$$v^2=2\ddot x_c l_0\;\rightarrow\; v^2=\ddot x_A l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+2m}}$$
It would be otherwise correct.

$$\ddot x_c=\frac{1}{2}\ddot x_A=\frac{4mgr^2}{2\left[ \frac{Mr^2}{2}+(M+4m)r^2 \right]}=\frac{4mg}{3M+8m}$$
$$v^2=2\ddot x_c l_0 \;\rightarrow \; v=2\sqrt{\frac{2mgl_0}{3M+8m}}$$
But still differs from the energy method in post #9 which gave:
$$v=2\sqrt{\frac{mgl_0}{M}}$$

Karol said:
$$-mgl_0+\frac{1}{2}mv^2=\frac{1}{2}I_B\omega^2$$
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.

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 )

dean barry said:
Ive assumed that linear distance moved by the hanging mass and the cylinder will be the same.
Not a good start.

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.

Karol said:
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.

ehild said:
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?

Karol said:
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}}##
Karol said:
Now i translate ω that i found into vc by ##v_c=\omega r##.
That is correct.

ehild said:
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.

Karol said:
$$v=2\omega r\;\rightarrow\; -mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
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.

Karol said:
$$-mgl_0+\frac{1}{2}m\cdot 4\omega^2r^2=\frac{1}{2}\left[ \frac{Mr^2}{2}+Mr^2 \right]\omega^2$$
$$Not good, negative under the square root. Your energy balance equation is wrong. The KE increases, the PE decreases. The total energy remains constant. ΔKE + ΔPE =0 What is ΔPE? What is the displacement of the hanging mass if the CM of the cylinder is displaced by lo?$$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 rmgl_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}}$$Like in the kinematic approach in post #11 Karol said:$$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 rmgl_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}}$$Like in the kinematic approach in post #11 Not quite. Compare them. There is still problem with the PE. Cylinder linear speed is half of the falling mass, ok : The acceleration of the falling mass = ( m * g ) / ( ( m * 0.375 ) + M ) The acceleration of the cylinder is half that value.$$-(-2mgl_0)=\frac{1}{2}mv^2+\frac{1}{2}I_B\omega^2,\; v=2\omega r2mgl_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}}

1. How does mass affect the speed of a falling object?

The mass of an object does not affect its speed when falling due to gravity. All objects, regardless of their mass, will fall at the same rate of 9.8 meters per second squared.

2. Can a cylinder be pulled without sliding if it has a mass attached to it?

Yes, a cylinder can be pulled without sliding if the force applied to pull the cylinder is greater than the force of friction between the cylinder and the surface it is resting on. The added mass attached to the cylinder can increase the force of friction, making it more difficult to pull without sliding.

3. How does the angle of the surface affect the pull force needed to move a cylinder without sliding?

The angle of the surface does not directly affect the pull force needed to move a cylinder without sliding. However, a steeper surface may increase the force of friction, making it more difficult to pull the cylinder without sliding.

4. Is there a difference in the pull force needed to move a cylinder without sliding on different surfaces?

Yes, the force of friction between the cylinder and the surface it is resting on can vary depending on the surface material. This can affect the amount of force needed to pull the cylinder without sliding.

5. Can the shape of the mass attached to the cylinder affect the pull force needed to move it without sliding?

Yes, the shape of the attached mass can affect the force of friction between the cylinder and the surface it is resting on. A larger or differently shaped mass may increase the force of friction, making it more difficult to pull the cylinder without sliding.

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