a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and rolls in a straight line without slipping. how much vertical height does it gain before it stops rolling

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xero099 xero099 · Answer 1 · 2020-04-09T03:49:59+02:00

Answer:

[tex]h = 0.0362\,m[/tex]

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

[tex]U_{g,A} + K_{A} = U_{g,B} + K_{B}[/tex]

[tex]U_{g,B} - U_{g,A} = K_{A} - K_{B}[/tex]

[tex]m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}[/tex]

The moment of inertia of the coin is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

After some algebraic handling, an expression for the maximum vertical height is derived:

[tex]m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}[/tex]

[tex]h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}[/tex]

[tex]h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]h = 0.0362\,m[/tex]

Cricetus Cricetus · Answer 2 · 2022-02-03T15:25:08+01:00

Given values:

By using Principle of energy conservation,

→ [tex]U_g, A+K_A = U_g,B+K_B[/tex]

→ [tex]U_g, B -U_g,A = K_A -K_B[/tex]

[tex]mgh = \frac{1}{2} I \omega_0^2[/tex]

and we know,

→ [tex]I = \frac{1}{2} mr^2[/tex]

Now,

→ [tex]mgh = \frac{1}{4} mr^2 \omega_0^2[/tex]

or,

→ [tex]h = \frac{r^2 \omega_0^2}{g}[/tex]

By putting the values,

[tex]= \frac{(0.0108)^2\times (55.2)^2}{9.807}[/tex]

[tex]= 0.0362 \ m[/tex]

Thus the response above is appropriate.

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