A solid ball (sphere) rolls without slipping along a horizontal surface, then down a ramp and then along a lower horizontal surface. The ball's center-of-mass velocity changes from u1 = 2.70 m/s to u2 = 4.50 m/s. What is the height, h, of the ramp?
HINTS: The height of the center-of-mass of the ball changes by a height equal to the height of the ramp. You will need to refer to Table 7.4 of the text for the moment of inertia of a solid sphere. As you work the problem, you will see that you do not need to know the mass or the radius of the ball.
HINTS: The height of the center-of-mass of the ball changes by a height equal to the height of the ramp. You will need to refer to Table 7.4 of the text for the moment of inertia of a solid sphere. As you work the problem, you will see that you do not need to know the mass or the radius of the ball.
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Thanks, chad, for referring me to Table 7.4, but I don't think I need it.
Use conservation of energy.
PE_top + KE_top = PE_bottom + KE_bottom
PE_top = mgh, where "h" is the ramp's height (the value we want to find). We aren't given "m", but don't worry about that--hopefully that variable will cancel out in the end.
KE_top is a combination of the balls translational KE (=½m(v_top)²), and its rotational KE (=½I(ω_top)²). That is:
KE_top = ½m(v_top)² + ½I(ω_top)²
"I" is the ball's moment of inertia, which equals (2/5)mR² (R=ball's radius). That was probably in that table 7.4 that you wanted me to look at. Again, we don't know "R", but don't fret it right now.
Since it's rolling without slipping, "ω_top", the ball's initial angular velocity, is related to its linear velocity like so:
ω_top = v_top/R
So that means the rotational KE is:
½I(ω_top)² = ½(2/5)mR²(v_top/R)² = ½(2/5)m(v_top)²
(Yay! the "R" cancelled out.)
And that means the total KE_top is
KE_top = ½m(v_top)² + ½(2/5)m(v_top)² = (7/10)m(v_top)²
Now at the bottom, we have:
PE_bottom = 0
And by an logic similar to the above:
KE_bottom = (7/10)m(v_bottom)²
Now, we said PE_top+KE_top = PE_bottom+KE_bottom. SO:
mgh + (7/10)m(v_top)² = (7/10)m(v_bottom)²
Divide by "m" and it magically disappears:
gh + (7/10)(v_top)² = (7/10)(v_bottom)²
Finally, just solve for "h" and plug in the values for v_top and v_bottom given in the problem.
Use conservation of energy.
PE_top + KE_top = PE_bottom + KE_bottom
PE_top = mgh, where "h" is the ramp's height (the value we want to find). We aren't given "m", but don't worry about that--hopefully that variable will cancel out in the end.
KE_top is a combination of the balls translational KE (=½m(v_top)²), and its rotational KE (=½I(ω_top)²). That is:
KE_top = ½m(v_top)² + ½I(ω_top)²
"I" is the ball's moment of inertia, which equals (2/5)mR² (R=ball's radius). That was probably in that table 7.4 that you wanted me to look at. Again, we don't know "R", but don't fret it right now.
Since it's rolling without slipping, "ω_top", the ball's initial angular velocity, is related to its linear velocity like so:
ω_top = v_top/R
So that means the rotational KE is:
½I(ω_top)² = ½(2/5)mR²(v_top/R)² = ½(2/5)m(v_top)²
(Yay! the "R" cancelled out.)
And that means the total KE_top is
KE_top = ½m(v_top)² + ½(2/5)m(v_top)² = (7/10)m(v_top)²
Now at the bottom, we have:
PE_bottom = 0
And by an logic similar to the above:
KE_bottom = (7/10)m(v_bottom)²
Now, we said PE_top+KE_top = PE_bottom+KE_bottom. SO:
mgh + (7/10)m(v_top)² = (7/10)m(v_bottom)²
Divide by "m" and it magically disappears:
gh + (7/10)(v_top)² = (7/10)(v_bottom)²
Finally, just solve for "h" and plug in the values for v_top and v_bottom given in the problem.