Three Cylinders (rotational kinematics)

In summary, three identical cylinders with solid, uniform density and mass of 16 kg and radius of 1 m are connected by a string and released from rest from a height of 4.4 m. Just before cylinder #3 hits the ground, it is moving at a speed of 5.36 m/s and cylinder #2 has a rotational speed of 5.36 rad/s. When cylinder #3 is replaced by a sphere of the same mass, it still has a speed of 5.36 m/s just before hitting the ground. However, when cylinder #1 is also replaced by a sphere of the same mass and radius, this time cylinder #3 is moving at a speed of 9.794 m/s just
  • #1
r34racer01
63
0
showmepl-2.gif


Three identical, solid, uniform density cylinders, each of mass 16 kg and radius 1 m, are mounted on frictionless axles that are attached to brackets of negligible mass. A string connects the brackets of cylinders #1 and #3 and passes without slipping over cylinder #2, whose bracket is attached to the ledge. Cylinder #1 rolls without slipping across the rough ledge as cylinder #3 falls downward.

This system is released from rest from the position shown -- with cylinder #3 at a height of 4.4 m above the ground.
a) How fast is cylinder #3 moving just before it hits the ground?
v = 5.36

b) What is the rotational speed of cylinder #2 at the time in (a)?
w = 5.36

c) Cylinder #3 is now replaced by a sphere of the same mass. How fast is it moving just before it hits the ground?
v'= 5.36

d) Finally, cylinder #1 is replaced by a sphere of the same mass and radius. How fast is cylinder #3 moving just before it hits the ground?
v'' = ?

This is where I'm having problems. This should be exactly the same as pt. A except we have a sphere so c=2/5. So I did
mgh = 2/5mv^2
(16)(9.81)(4.4) = 2/5(16)v^2
and I got 10.388 = v'' but apparently that's wrong, anyone know what's going on?


 
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  • #2
r34racer01 said:
So I did
mgh = 2/5mv^2
(16)(9.81)(4.4) = 2/5(16)v^2
and I got 10.388 = v'' but apparently that's wrong, anyone know what's going on?

Hi r34racer01! :smile:

You've left out the energy of cylinder #2.
 
  • #3
tiny-tim said:
Hi r34racer01! :smile:

You've left out the energy of cylinder #2.

Ok now I tried 690.624 = (1/2)*I(sphere)*w^2 + (1/2)*I(cylinder)*w^2
690.624 = (1/2)((2/5)mr^2)(v^2/r^2) + (1/2)((1/2)mr^2)(v^2/r^2)
690.624 = (0.2)mv^2 + (0.25)mv^2
690.624 = 3.2v^2 + 4v^2
9.79388 = v
But that's still wrong. What am I doing wrong?
 
  • #4
sorry!

Hi r34racer01! :smile:

I'm sorry I've taken so long to reply :redface:

(have an omega: ω and a squared: ² :smile:)
r34racer01 said:
Ok now I tried 690.624 = (1/2)*I(sphere)*w^2 + (1/2)*I(cylinder)*w^2
690.624 = (1/2)((2/5)mr^2)(v^2/r^2) + (1/2)((1/2)mr^2)(v^2/r^2)
690.624 = (0.2)mv^2 + (0.25)mv^2
690.624 = 3.2v^2 + 4v^2
9.79388 = v
But that's still wrong. What am I doing wrong?

You've left out the translational KE of masses 1 and 3. :smile:
 

FAQ: Three Cylinders (rotational kinematics)

What is rotational kinematics?

Rotational kinematics is a branch of physics that deals with the motion of objects that are rotating or spinning around a fixed axis.

What are the three cylinders in rotational kinematics?

The three cylinders in rotational kinematics refer to the three principal axes of rotation: the x-axis, y-axis, and z-axis.

How do you calculate the angular velocity of a rotating object?

The angular velocity of a rotating object can be calculated by dividing the change in angular displacement by the change in time. This is represented by the equation ω = Δθ/Δt, where ω is the angular velocity in radians per second, Δθ is the change in angular displacement in radians, and Δt is the change in time in seconds.

What is the difference between angular velocity and linear velocity?

Angular velocity is a measure of how fast an object is rotating, while linear velocity is a measure of how fast an object is moving in a straight line. Angular velocity is measured in radians per second, while linear velocity is measured in meters per second.

What is the relationship between angular velocity and angular acceleration?

Angular velocity and angular acceleration are related by the equation ω = ω0 + αt, where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time interval. This equation shows that the final angular velocity is equal to the sum of the initial angular velocity and the product of the angular acceleration and time.

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