Question

(a) A sphere and a cylinder are released from the top of an inclined plane and roll without slipping to the bottom. Which one gets there first? (explain your answer and show calculations) [2 Marks] (b) What does it mean to say that an object is dynamically balanced? How is this different to an object being statically balanced? [1 Mark] (c) Under what conditions can we apply techniques from 2-dimensional (planar) rigid body mechanics to 3-dimensional objects? [1 Mark] (d) A 500g superball collides with a 2kg slender rod as shown below; 0.4m 3ms 1.5m Assuming the collision conserves energy; Calculate the velocity of the ball and rod (centre of gravity), and the angular velocity of the rod after the collision has occurred [6 Marks] Ignore the effects of gravity and consider the superball as being a point mass.

Answer 1

a) The moment of inertia for a solid cylinder is I = (¹/₂) m r² and for a solid sphere is I = (²/₅) m r².

he total energy of each object is given as

E=mgh+1/2 mv² +1/2 Iω²

subsituting v=ωr (for no slip condition) and the MI for sphere and cylinder the energy for the sphere and cylinder becomes

Es=mgh+7/10 mv² for the sphere and Ec=mgh+3/4mv² for the cylinder.

As the total energy must remain constant and equal for both objects , for each height h the sphere must have a higher velocity. Therefore, the sphere will reache the bottom first.

b) An object is said to be dynamically balanced, if it can rotate without producing and resultant force or couple and withut the need of any supporting counterbalancing force other than to support its own weight.

An object is aid t be statically balanced if all the forces and reactions cancel each other out. In this case the object remains stationary.

c) If the thickness of the object is uniform in the 3rd dimension and the forces and reactions are acting in the other two directions only, 2D techniques can be applied to 3D objects

d) by law of conservation of momentum

0.. * 3 + 2 * 0 = 0.5 * vit202

1.5 0.5v1 +2v2

since energy is conserved in in the collision

0.5*32 +2 * 0 0.5 * +21,2

on solving (i) and (ii) we get

v1--1.8 m/s and v2- 1.2 m/s

, also for the rod

w = v/r = 1.2/0.75 1.6 rad/s