Solve Rotational Motion w/ Pulley & Cylinder: Acceleration Calculation

In summary, the block is falling and the cylinder is rotating. The cylinder has a mass and is falling, so it has an inertial force, which is the tension in the string. The block has a mass and is falling, so it has a gravitational force. The sum of the forces is 1.5m1R2. a is .5m1R2.
  • #1
SuperGeek
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0
I am stuck on this rotational motion with pulley question. Any help would be appreciated.

A block M2=5kg hangs off a pulley with mass from a string with tension T2 over a table edge.
A cylinder M1=3kg on the horizontal table top is attached to the string on the other end of the pulley (there is friction between the cylinder and table).
The cylinder is allowed to roll while the block is allowed to fall.
I need to compute linear acceleration of the roller-pulley-block system. The problem I have is that I am not sure what to use as the horizontal force on the cylinder.

Forces on m2: M2g - T2 = M2a
Forces on pulley: T2-T1 = 1\2ma
Forces on m1: N - m1g = 0 (vertical axis)
T1 - f = m1a
Should f = (Inertia of center of mass \ R^2)(a) ?

Thanks in advance for any responses.
 
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  • #2
I am little confused about your problem. The pulley with mass is also a cylinder? Since it has mass, it must rotate.
 
  • #3
Unless you've left something out of the problem statement, it doesn't seem like you have to consider inertial effects of the pulley (I certainly doubt the pulley is moving, as your equations seem to suggest). If you want to use a Newtonian approach (as opposed to an energy-based method) as you have started, you're first equation is fine:
[itex] m_2a = m_2g - T [/itex]

where T is the tension in the string and you have taken a>0 to be a downward acceleration of m2.

Assuming the string is inextensible, there is only one acceleration to deal with in this problem. Also, with the assumption of a massless, frictionless pulley, you can say that the tension on either end of the pulley is the same. This gives your x-direction equation for the cylinder to be:
[itex] m_1a = T - f [/itex]

You still need another equation, so look at a torque balance for the cylinder. (Another assumption here: the string loops around a groove or bearing in the cylinder (like a yo-yo, maybe) so that the cylinder can roll without slip while being pulled; there's no unwinding string to worry about.) Look at the sum of torques about the center of the cylinder:
[itex] I\alpha = fR [/itex]
But we also know how angular acceleration relates to linear acceleration:
[itex] \alpha = \frac{a}{R} [/itex]
(watch your sign conventions; I believe what I've typed above is consistent...)

Now you can put it all together, along with I = .5*m1R2, and solve for a.
 

1. How do I calculate the acceleration in rotational motion with a pulley and cylinder?

To calculate the acceleration in rotational motion with a pulley and cylinder, you will need to use the formula a = (r * α), where a is the acceleration, r is the radius of the cylinder, and α is the angular acceleration. You will also need to consider the mass and moment of inertia of the objects involved.

2. What is the relationship between the radius of the cylinder and the acceleration?

The radius of the cylinder has a direct relationship with the acceleration in rotational motion. This means that as the radius increases, the acceleration also increases. This is because the greater the radius, the more distance the object has to travel in a given period of time.

3. Can you use the same formula for calculating acceleration in both clockwise and counterclockwise rotation?

Yes, the formula for calculating acceleration in rotational motion with a pulley and cylinder can be used for both clockwise and counterclockwise rotation. The only difference is that the direction of the angular acceleration will be positive or negative depending on the direction of rotation.

4. How do you take into account the effect of friction in this type of rotational motion?

To take into account the effect of friction in rotational motion with a pulley and cylinder, you will need to add the frictional torque to the equation. This can be calculated using the formula τ = μ * r * F, where τ is the frictional torque, μ is the coefficient of friction, r is the radius, and F is the force acting on the object.

5. Are there any other factors that can affect the acceleration in this type of rotational motion?

Yes, there are other factors that can affect the acceleration in rotational motion with a pulley and cylinder. These include the mass of the objects involved, the applied force or torque, and any external forces such as air resistance. It is important to consider all of these factors when calculating the acceleration in rotational motion.

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