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## Homework Statement

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A horizontal force, ##F##, is applied to a cylinder of mass ##M## and radius ##R## (the line of action of ##F## goes through the center) in order for the cylinder to climb over a step of height ##R/2##.

1) Find the torque around ##A## and find force ##F## as a function of the normal force ##N## with the floor for the system to be in equilibrium. (Note that the minimum force needed to climb up the step is obtained when ##N##=0).

2) Write down the forces once the cylinder has rotated an angle ##\theta## and has left the floor. Which forces do work?

3) The cylinder finishes its ascent once it is completely over the height of the step. At that moment, force ##F## stops being applied. What is the rotational velocity of the cylinder? What is the velocity of the center of mass? (both as functions of ##F##). Consider ##I_{CenterOfMass}## given.

## Homework Equations

Choosing axes from A: x positive towards the right, y positive upwards, rotation positive counterclockwise. Calling the normal force from the step ##A## and the normal force from the floor ##N##.

If I draw a right triangle between the center of the cylinder and A, the hypothenuse is R, the shorter leg is R/2, that makes the longer leg ##R\ \sqrt{3}/2##. Let's call that ##d##.

**##\sum\ \ F_x = F-A_x = m \ddot x ##**

1)

1)

##\sum\ \ F_y = A_y - mg + N = m \ddot y ##

##\sum\ \tau_A = -FR/2 + mgd - Nd = I_A \dot \omega = (I_{CM} + mR^2) \dot \omega##

**##\sum\ \hat \theta = - mg \ cos \ \theta + F \ sin \ \theta + A \ cos \ \theta = m r \ddot \theta ##**

2)

2)

##\sum\ \hat r = - mg \ sin \ \theta - F \ cos \ \theta + A \sin \ \theta= -m r \dot \theta^2 ##

##W= \vec F \cdot \vec s##

## The Attempt at a Solution

1) When the system is at equilibrium, ##\ddot x = 0, \ddot y = 0, \dot \omega = 0##

##\vec F=(mg-N)d/R \ \ \hat x =>## ## \hat F = (mg-N)\ \sqrt{3} \ \hat x##

**##W= \vec F \cdot \vec s## The weight and F do work.**

2)

2)

**I'm still working on the third part, but I am confused about the direction of force A. If I calculate it from the first part of the exercise, when the cylinder is still touching the ground:**

##A_x = (mg-N)\ \sqrt{3}## and ##A_y = (mg-N)## therefore the direction of the force is ##tan \alpha = 1/ \sqrt{3}##, ## \alpha= \pi/6##

Is there anything that should be telling me that the force A becomes purely radial after the cylinder leaves the floor or should I carry on with ##\hat \theta## and ##\hat r## ?

Is there anything that should be telling me that the force A becomes purely radial after the cylinder leaves the floor or should I carry on with ##\hat \theta## and ##\hat r## ?

Thanks!