Question

Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. In which order do the objects reach the bottom of the incline?a) 1, 2, 3 b) 2, 3, 1 c) 3, 1, 2 d) 3, 2, 1 e) All three reach the bottom at the same time.

Answer 1

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = $\frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}$

$v_{Sphere}$ = $\sqrt{\frac{10}{7} gh}$

$v_{Hoop}$ = $\sqrt{gh}$

$v_{Disc}$=$\sqrt{\frac{4}{3} gh}$

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3