Physics problem with cylinder.?

Question

Hello,

First, here is the drawing which shows the situation:

http://img409.imageshack.us/my.php?image=33yb6.jpg

So there is a cored cylinder which radius is R and angular velocity is omega. The cylinder is placed in the the corner so the cylinder touches the wall and the ground. Coefficient of friction between the wall and the cylinder is equal to the coefficient of friction between the ground and the cylinder (this friction coefficient is equal to μ). After how many turns the cylinder will stop to rotate?

Thanks in advance.

So what is the right solution of the problem?

Answer 1

Edward:

The force from the floor μmg results in extra μμmg from the wall.
That extra μμmg force from the wall results in extra-extra force μμμmg from the floor.
That extra-extra μμμmg force from the floor results in extra-extra-extra force μμμμmg from the wall....


***************************
Draw force balance diagram.
There are three forces:

gravity mg,
F1 acting from the floor
F2 acting from the wall

You now have right angle triangle with angle arctan(μ) and hypotenuse mg. When there is sliding, net force of reaction (normal + friction) is always inclined at angle arctan(μ) to the perpendicular to the surface.

Torques of the forces in clockwise direction are respectively
0,
μmgR/(1+μ²),
μ²mgR/(1+μ²).

Net torque is
T = μ(μ+1)/(1+μ²) mgR

Answer 2

Total force
F=f1 + f2
Force acting on slowing the cylinder down is the sum of floor/cylinder friction force and the wall/cylinder friction force
F= μmg(1+μ)

The cylinder has kinetic energy
E= 0.5 I w^2
where I= mr^2 moment of inertia
w - angular velocity

The work done by friction to stop the cylinder equal to the E of the cylinder and work done by F.
W=E
W= twrF= 0.5 mr^2 w^2

number of turns N=wt then
wt= 0.5 mr^2 w^2/(F r)
N= 0.5 m r w^2/ (μmg(1+μ)) finally
N= 0.5 r w^2/ (μg(1+μ))

Alexander:

As always you have a good point! However today...
...you got the direction mixed up. The cylinder will climb the wall and therefore reduce the normal force on the floor, which in turn will drop not increase the normal force on the wall and that will force the cylinder even to slip even more freely.

Depending on μ and the relationship of static and dynamic friction we will actually have a nice damped osculation and with μ=0 it will never stop.
(Please note that to make this problem more realistic we need both static and dynamic μ. )


Pythagor,
You have presented us with a good problem! I have provided you with a first order approximation.

Answer 3

a million. First we would desire to locate how long the cannonball is indoors the air, its "hang time" in case you will. This span of time is the time it takes for the cannonball to pass up, and fall circulate into opposite. it somewhat is thoroughly based on the y-portion of the fee, so we would desire to apply some trigonometry to locate this. in case you drew a staggering triangle, you will stumble on that the y-comp onent: velocity at y axis = (preliminary velocity)sin(seventy 9) = 993sin79 = 974.756 m/s. 2. Now which you have the fee on the y axis, we are waiting to now stumble on the time it takes for the cannon to land. we are waiting to observe the equation very very final velocity = preliminary velocity + (Acceleration)(Time) or Vf = Vi + at observe that your aceeleration = -g, because of the fact's pulling it downward so: Vf = Vi - gt. putting aside t: Vf + gt = Vi gt = Vi - Vf t = (Vi - Vf)/g all and sundry understand the preliminary velocity indoors the y axis, it somewhat is 974.756 m/s. yet what with regard to the recommendations-blowing velocity? good right this is something interesting: as quickly as you throw a ball up at say a diverse velocity 2m/s, at the same time because it falls lower back on your hand (assuming you saved your hand on the comparable place) it is going to return on your hand at a velocity of -2m/s, the call of the preliminary velocity. So contained at the same time because it includes the cannonball, i visit assert that the recommendations-blowing velocity is -974.756 m/s. i understand my Vi, Vf, and g, so i visit now therapy for time. t = (974.756 - (-974.756))/9.8 = (974.756 + 974.756)/9.8 = 1949.512/9.8 = 198.9298 seconds. 3. Why do i'd desire to admired the hang time? it somewhat is thru utilising fact the ball is traveling on the x direction for this lots time, and after that element, it stops shifting (till ultimately it keeps on rolling, yet it somewhat is yet yet another situation! enable's assume it maintains to be placed the area it lands.) besides, i visit apply the equation velocity = distance/time or distance = (velocity)(time) = vt I certainly have my time, yet i p.c. my velocity. submit to in recommendations we are looking for the section on the x-direction, now not the y, so we for the fee good right here, we would desire to get the x-portion of the projectile's velocity. using trigonometry lower back: velocity at x axis = (preliminary velocity)cos(seventy 9) = 993cos79 = 993(0.19) = 189.473 m/s. So, using this on the previous equation, and the air-time: distance traveled alongside the horizontal direction = vt = (189.473)(198.9298) = 37,691.826 meters.

Answer 4

1 turn