Rotational Dynamics - In the scenario that a ball is rolled down an inclined pla

Question

Rotational Dynamics - In the scenario that a ball is rolled down an inclined plane, and then at the end of the track it falls to the ground off of the table in which the inclined plane is resting, what would happen in these alternate scenarios? Here is an image of the proposed scenario pathway. h is the height of the rod that is holding the inclined plane at a 15 degree angle (h = 31.12 cm). y is the height of the table (60.96 cm). x is the distance from the table to the fall point. The x value changes with different objects (gold disk, silver disk, and sphere).

1) If the ball rolled down were twice as heavy(i.e., had twice the mass) , would the distance between the table and landing point change? Why?

2)If the ball you rolled down the ramp were twice as large (i.e., had twice the radius), how would this affect your results?

3)If you rolled a cylinder down the ramp, what would be its speed at the bottom? (Let the vertical distance, h, be 6.35 cm. The moment of inertia of a cylinder is the same as that of a disk.)

4) If you rolled a ring down the ramp, what would be its speed at the bottom? (Let the vertical distance, h, be 6.35 cm.)

5)If you rolled the ball, cylinder and ring at the same time, in what order would they reach the bottom?

6)If you dropped the ball, cylinder and the ring at the same time from a height of 6.35 cm, in what order would the hit the ground?

Explanation / Answer

(1) The landing distance depend upon the speed that the ball has at the bottom of incline plane.
When mass will change the velocity will remain unchange at the bottom , we can see it from the following expression.
Initial energy = (mgh)
Final energy = (1/2)mv2 + (1/2)Iw2
where m is mass of the ball , v is the velocity at the bottom , I is moment of inertia and w is angular velocity.
I = (2/5)mr2 and w = v/r
Putting these values
Final energy = (1/2)mv2 + (1/2)(2/5)mr2 *(v/r)2 = (1/2)mv2+ (1/5)mv2 = (7/10)mv2
Now applying energy conservation
intial energy = final energy
mgh = (7/10)mv2
v2 = (10/7)gh
Hence velocity is independent of mass, therefore its landing distance will remain unchanged on increasing the mass.
(2) As we can see from the above expression that velocity is independent of radius also, therefore
on increasing the radius also couldn't affect the landing distance.
(3) For a cylinder we know that the moment of inertia (I) = mr2 /2
So final energy = (1/2)mv2 + (1/2)Iw2 = (1/2)mv2 + (1/2)(mr2 /2)(v/r)2 = (1/2)mv2 +(1/4)mv2 = (3/4)mv2
Initial energy = mgh
so applying energy conservation
mgh = (3/4)mv2
9.81*(6.35*10-2) = (3/4)*v2
v = 0.911 m/s
(4) For a ring I = mr2
Final enery = (1/2)mv2 + (1/2)*(mr2)(v/r)2 = (1/2)mv2 + (1/2)mv2 = mv2
So according to energy conservation
mgh = mv2
v2 = gh = 9.81*(6.35*10-2)
v = 0.789 m/s

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