Question

A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. The solid sphere was released from a height of 19.3 cm. From what height hcylinder should the solid cylinder be released so that it has the same speed as the solid sphere when it reaches the bottom of the ramp?

Answer 1

Answer:

Height = 20.6 cm

Explanation:

We can solve this problem by using the conservation of mechanical energy.

In fact, for each of the two objects, the initial gravitational potential energy when it is at the top of the ramp is entirely converted into translational kinetic energy + rotational kinetic energy. So we can write for each object

Solid Sphere

KE1 + PE1 = KE2 + PE2

$0 + m g h = \frac{1}{2} m v^2 + \frac{1}{2} I w^2 + 0\\\\I = \frac{2}{5} m r^2\\\\m g h = 0.5 m v^2 + 0.5 * 0.4 m r^2 (v/r)^2\\\\m g h = 0.5 m v^2 + 0.2 m v^2$

$g h = 0.7 v^2\\\\v = \sqrt{(gh/0.7)} \\\\= \sqrt{( 9.81 \times 0.193 / 0.7)} \\\\= 1.645 m/s$

Solid Cylinder

KE1 + PE1 = KE2 + PE2

$0 + m g h = \frac{1}{2} m v^2 + \frac{1}{2} I w^2 + 0$

$I = \frac{1}{2} m r^2$

$m g h = \frac{1}{2} m v^2 + \frac{1}{2} * \frac{1}{2} m r^2 (v/r)^2$

$m g h = 0.5 m v^2 + 0.25 m v^2$

$g h = 0.75 v^2$

$h =\frac{ 0.75 v^2}{g} \\\\=\frac{ 0.75 \times (1.645)^2}{ 9.81} \\\\= 0.206 m \\\\= 20.6 cm$

Height = 20.6 cm

Answer 2

Answer:

H = 14.25cm

Explanation: