A marble spins in a vertical plane around the inside of a smooth, 20-cm-diameter

Question

A marble spins in a vertical plane around the inside of a smooth, 20-cm-diameter horizontal pipe. The marble's speed at the bottom of the circle is 3.0 m/s.

A) What is the marble's speed at the top?

B) The marble's position in the pipe can be specified by an angle theta measured counterclockwise from the bottom of the pipe. Find an algebraic expression for the marble's speed when it is at angle theta. Use numerical values for R, g, and the initial speed, leaving theta as the only symbol in the equation. Your expression should give 3.0 m/s for theta = 0degrees and your answer to part A for theta = 180degrees.

Explanation / Answer

At the top, it needs to have centripetal acceleration of v^2/r>g So is v^2/r greater than g? Now its speed at the top is determined by the bottom speed minus losses to gravity. speed at top^2 =speedbottom^2-2g*.20 find the square root of both sides and then find speed at top the 2g comes about because its diametre /2 ans= 2.236m/s as for theta = v^2 = 3m.s^2- 10^2/r sin (theta/2) take g = 10m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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