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
[tex]KE_r = \dfrac{1}{2}I\omega^2[/tex]
Moment of inertia of coin
[tex]I = \dfrac{1}{2}MR^2[/tex]
so,
[tex]KE_r = \dfrac{1}{4}MR^2\omega^2[/tex]
Transnational Kinetic energy
[tex]KE_t = \dfrac{1}{2}Mv^2[/tex]
v = r ω
[tex]KE_t = \dfrac{1}{2}MR^2\omega^2[/tex]
now,
using conservation energy
Kinetic energy of the coin is converted into the potential energy
KE_r + KE_t = PE
[tex] \dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh[/tex]
[tex]\dfrac{3}{4}R^2\omega^2=gh[/tex]
[tex]\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h[/tex]
h = 0.0259 m
Vertical height gain by the coin is equal to 0.0259 m
