Linear and Rotational Kinematics System.

In summary, the block of mass 3.51724 m/ss falls faster than the block of mass 2.83636 m/ss due to the fact that the block of mass 3.51724 m/ss has a greater coefficient of friction with the table.
  • #1
CaptainSFS
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0

Homework Statement


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A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.53. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?

Hint: For each block, apply Newton's 2nd law. For the pulley, apply the rotational analog of Newton's 2nd law. And remember that since the string does not slip on the pulley, there is a relationship between the acceleration of the blocks and the angular acceleration of the pulley. You will end up with three equations and three unknowns (the answers to a,b,c). EDIT: is pulley -> (1/2)m*r*a = T3-T1?

Homework Equations



Kinematics

The Attempt at a Solution



I believe I created the correct formulas for each mass.

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: Torque = I*alpha -> Torque = (1/2)m2*r*a

I'm however not sure if these are correct, but more importantly, how I would combine them in order to find the acceleration of the falling mass.

I realize that I need 3 equations and unknowns (T1, T2, and a). I believe my Pulley isn't correct because it also has T. Is my assumption correct? And, How would I change the pulley equation so that it works?

EDIT: I believe now that my pulley should be, (1/2)m*r*a=T3 - T1. Is this correct?

EDIT: Still having trouble. Using my above assumptions I came up with this: (m3*g-m1*g*u)/(m3+m1+.5*m2*r)=a
This unfortunately still doesn't work. Any help would be much appreciated, I'm really trying my hardest to figure this out. :)
 
Last edited:
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  • #2
Hi CaptainSFS,

CaptainSFS said:

The Attempt at a Solution



I believe I created the correct formulas for each mass.

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: Torque = I*alpha -> Torque = (1/2)m2*r*a

I'm however not sure if these are correct, but more importantly, how I would combine them in order to find the acceleration of the falling mass.

I realize that I need 3 equations and unknowns (T1, T2, and a). I believe my Pulley isn't correct because it also has T. Is my assumption correct? And, How would I change the pulley equation so that it works?

EDIT: I believe now that my pulley should be, (1/2)m*r*a=T3 - T1. Is this correct?

I don't believe this is quite right. The right hand side (T3-T1) is not the torque. What would the torque be? (Think about what the units of torque are.)
 
  • #3
torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?

EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/
 
Last edited:
  • #4
CaptainSFS said:
torque has units of Nm. So I need some sort of distance I take it? Would I use the radius as this distance? Would it be Torque=(T3-T1)r ?

Yes, that's right.

EDIT: This also doesn't seem to work. I'm feeling pretty lost at this point. =/

Can you show more of your work? What numbers are you using, and what is your final answer?
 
  • #5
Alright

Block 1: T1 - Frictional Force = m*a

Block 3: m3*g - T3 = m*a

Pulley 2: (1/2)m2*r*a = (T3 - T1)*r

Then, T1 = m1*a + m1*g*u | T3 = m3*g - m3*a <--I plug these two into my 'Pulley 2'

I then algebraically solve for a and I end up with,

a = (m3*g - m1*g*u) / (0.5*m2 + m1 + m3)

I use m1=2kg m2=0.4kg u=.53 m3 = 4kg g=9.81m/ss || The answer I get is 3.51724 m/ss

EDIT: Okay, So I plugged it in again, and I got 4.6518m/ss, which is the right answer. :P

Thanks for your help! :)
 
  • #6
Glad to help!
 

1. What is the difference between linear and rotational motion?

Linear motion is when an object moves in a straight line, while rotational motion is when an object spins or rotates around an axis. Linear motion can be described by displacement, velocity, and acceleration, while rotational motion can be described by angular displacement, angular velocity, and angular acceleration.

2. How do you calculate the velocity of an object in a rotational motion?

The velocity of an object in rotational motion can be calculated by dividing the angular displacement by the time taken, or by multiplying the angular speed by the radius of rotation. The unit of angular velocity is radians per second (rad/s).

3. What is the relationship between linear and angular velocity?

The relationship between linear and angular velocity is that they are directly proportional to each other. This means that if the linear velocity of an object increases, the angular velocity will also increase, and vice versa.

4. How is rotational motion affected by forces?

In rotational motion, forces can cause an object to rotate around an axis. The direction of the force will determine the direction of the rotation, while the magnitude of the force will determine the angular acceleration of the object.

5. What are some real-life examples of linear and rotational motion?

Examples of linear motion include a car moving in a straight line, a person walking in a straight line, or a ball rolling down a hill. Examples of rotational motion include a spinning top, a merry-go-round, or a figure skater spinning on the ice.

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