This is an algebraic problem. No numbers are involved. How high does the marble

Question

This is an algebraic problem. No numbers are involved.

How high does the marble of radius, rand mass m have to start to make it around the loop of radius R?

Assume the marble is a solid sphere with moment of inertia around a diameter:  

You may also assume the marble does not slip as it rolls.

How high does the marble of radius, r and mass m have to start to make it around the loop of radius R? Assume the marble is a solid sphere with moment of inertia around a diameter: You may also assume the marble does not slip as it rolls.

Explanation / Answer

Moment of inertia of solid sphere= 2/5 mr^2

for this problem go backwards from top loop to bottom of loop,

energy needed at the bottom of loop will be KE and to reach it to top it must have potential energy equal to mg(2R)

So,

KE at bottom = mg(2R)

KE at bottom = .5Iw^2

.5Iw^2 = mg(2R)

to reach this KE, let it start from height H

so:

mgH = .5Iw^2

so,

mgH = mg(2R)

H = 2R

negleting all loses

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.