Question

A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls in a straight line without slipping. How much vertical height does it gain before it stops rolling? The acceleration due to gravity is 9.81 m/s 2 . Answer in units of m.

Answer 1

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

                     radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

$KE_r = \dfrac{1}{2}I\omega^2$

Moment of inertia of coin

$I = \dfrac{1}{2}MR^2$

so,

$KE_r = \dfrac{1}{4}MR^2\omega^2$

Transnational Kinetic energy

$KE_t = \dfrac{1}{2}Mv^2$

v = r ω

$KE_t = \dfrac{1}{2}MR^2\omega^2$

now,

using conservation energy

Kinetic energy of the coin is converted into the potential energy  

KE_r + KE_t = PE

$\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh$

$\dfrac{3}{4}R^2\omega^2=gh$

$\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h$

h = 0.0259 m

Vertical height gain by the coin is equal to 0.0259 m