A uniform cylinder of mass M and radius R is launched up an inclined plane which

Question

A uniform cylinder of mass M and radius R is launched up an inclined plane which

makes and angle ? to the horizontal as shown below. The cylinder rolls without

slipping and starts up the incline with initial velocity VCM (CM meaning center of mass).

a) What is the direction of the angular velocity ?, clockwise or counter-clockwise?

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A uniform cylinder of mass M and radius R is launched up an inclined plane which makes and angle ? to the horizontal as shown below. The cylinder rolls without slipping and starts up the incline with initial velocity VCM (CM meaning center of mass). What is the direction of the angular velocity ?, clockwise or counter-clockwise? What is the direction of the angular acceleration ?, clockwise or counter-clockwise? What distance d up the incline does the cylinder travel before it comes to a stop? What is the linear acceleration of the cylinder as it is traveling up the incline? There is a force of static friction fs between the cylinder and the surface of the incline, with direction as shown in the figure. Determine the magnitude of fs.

Explanation / Answer

a) As it is rolling w/o slipping... so .. direction of angular velocity = clockwise

b) As the cylinder will move up, the angular velocity will decrease... so direction
of angular acceleration = counter-clockwise

c)

Initially, total energy = translational kinetic energy + rotational kinetic energy

translational kinetic energy = (1/2) * M * V_CM^2
inertia of cylinder = M * R^2 / 2
V_CM = w*R
so angular velocity w = V_CM / R

so rotational kinetic energy = 0.5 * inertia * w^2 = 0.5 * M*R^2 /2 * V_CM^2 / R^2 = (1/4) * M*V_CM^2

so total kinetic energy = (3/4) * M * V_CM^2 J


when the cylinder comes to rest.. all this kinetic energy will be converted to potential energy..

the height moved up the incline = d.
so vertical height = d*sin theeta..

so.. M * g * dsin theeta = (3/4) * M * V_CM^2

so . d = (3/4) * V_CM^2 / ( g sin theeta )


d) initial velocity = V_CM

final velocity = 0
distance = (3/4) * V_CM^2 / ( g sin theeta )

v^2 = u^2 + 2*a*d
0 = V_CM^2 + 2*a*(3/4) * V_CM^2 / ( g sin theeta )

so.. acceleration = a = - ( (2/3) * g sin theeta )
i.e. linear accelearion = ( (2/3) * g sin theeta ) down the incline


e) angular acceration = linear acc / radius = (2/3) * g sin theeta / R
so.. fs * R = inertia of cylinder * angular accelearion = (1/2)* M*R^2 * (2/3) * g sin theeta / R

so.. fs = (1/3) * g sin theeta

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