Answer:
This question can be answered using the conservation of energy.
[tex]K_1_{rot} + K_1_{trans} + U_1 = K_2_{rot} + K_2_{trans} + U_2\\\frac{1}{2}I\omega_1^2 + \frac{1}{2}mv_1^2 + 0 = \frac{1}{2}I\omega_2^2 + \frac{1}{2}mv_2^2 + mgh\\v = \omega R\\\frac{1}{2}\frac{2}{5}mR^2 (\frac{v_1}{R})^2 + \frac{1}{2}mv_1^2 = \frac{1}{2}\frac{2}{5}mR^2(\frac{v_2}{R})^2 + \frac{1}{2}mv_2^2 + mg(3\sin(25^\circ))\\\frac{1}{5}mv_1^2 + \frac{1}{2}mv_1^2 = \frac{1}{5}mv_2^2 + \frac{1}{2}mv_2^2 + mg(1.26)\\\frac{7}{10}v_1^2 = \frac{7}{10}v_2^2+ (9.8)(1.26)\\[/tex]
[tex]0.7(5.5)^2 = 0.7v_2^2 + 12.35\\0.7v_2^2 = 21.175 - 12.35 = 8.825\\v_2^2 = 12.6\\v_2 = 3.55~m/s[/tex]
Explanation:
The moment of inertia of a solid sphere is
[tex]I_{sphere} = \frac{2}{5}mr^2[/tex]
The potential energy is found by using the vertical distance, not the 3.00 m up the ramp. We multiply this 3.00 m by sine of 25° to find the vertical distance.
In rolling without slipping the translational speed is related to rotational speed through [tex]v = \omega R[/tex].
The speed of the sphere after it has rolled 3.00 m up the ramp is mathematically given as
v_2 = 3.55 m/s
Question Parameters:
a horizontal surface with a speed of 5.50 m/s
a ramp that makes an angle of 25.0°
Generally, the equation for the conservation of energy is mathematically given as
[tex]0.5w_1^2 + \frac{1}{2}mv_1^2 + 0 = \frac{1}{2}Iw_2^2 + \frac{1}{2}mv_2^2 + mgh[/tex]
Where
v=wR
[tex]\frac{1}{5}mv_1^2 + \frac{1}{2}mv_1^2 = \frac{1}{5}mv_2^2 + \frac{1}{2}mv_2^2 + mg(1.26)[/tex]
[tex]\frac{7}{10}v_1^2 = \frac{7}{10}v_2^2+ (9.8)(1.26)[/tex]
[tex]0.7(5.5)^2 = 0.7v_2^2 + 12.35[/tex]
v_2 = 3.55 m/s
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