Well, I thought of the question myself:
There are two identical straight slopes, 300 to the horizontal and 10m long at their hypotenuse. Each of them has one cylinder of identical shape of r=1m and mass (1kg) respectively. However, one is solid whereas the other is hollow.
They are attracted downwards by the same gravity force (10N, suppose gravitational accerleration is 10ms-2), and as a result of the support
force perpendicular to the slope's surface, an identical pull force parallel to the surface is pulling each cylinder (5N, derived from: sin300=F/Fg).
Deduce the time taken for each cylinder to reach the bottom of the slopes!
(Of course, the hollow will take a longer time because it has a greater rotational inertia due to its mass concentrated on further distance compared to the solid cylinder, hence angular acceleration is bigger)
Rotational inertia of hollow cylinder = mr2
Rotational inertia of solid cylinder = mr2/2
Torque = Rotational inertia*angular acceleration
Torque = force*perpendicular distance
linear acceleration = angular acceleration*radius
linear distance = initial linear speed*time taken + linear acceleration*time taken2/2
(initial linear speed = 0)
3. The attempt at the solution
When i tried to solve this question, i cannot do it without calculating or assuming friction force between the cylinders and the slope's surface. (Hence, the force of 5N which pulls the cylinders is not needed).
Suppose, the friction is 1N, hence torque is 1Nm.
Angular acceleration of the solid cylinder is Torque/rotational inertia = 1/0.5 = 2radians/s
Angular acceleration of the hollow cylinder is Torque/rotational inertia = 1/1 = 1radians/s
Linear acceleration of the solid cylinder = 2ms-2
Linear acceleration of the hollow cylinder = 1ms-2
hence using the last equation and rearrangements, time taken for solid cylinder is 101/2s
and time taken for hollow cylinder is 201/2s
Am i right?
can the problem be solved without assuming/calculating friction?