A sphere (with mass M, radius R and moment of interia of I =2/5MR^2 about its ce

Question

A sphere (with mass M, radius R and moment of interia of I =2/5MR^2 about its center) is at rest at the top of an inclined plane (with an angle from the horizontal of theta). It is released and begins to roll without slipping down the plane a distance d. Find an expression for the tangential speed of the sphere when it is a distance d down the slope from where it started. Your expression may include M, R, theta, d, g and numbers but nothing else. If M = 10 kg, R = 0.3 m, theta = 10 degrees, and d = 2 m, and h = 5 m, what is the speed of the sphere when it is a distance d from where is started? Include units all the way through your calculations!

Explanation / Answer

Part A)

At a distance d, it has traveled a height h where h = dsin(angle)

Then, by conservation of energy mgh = .5mv^2 + .5Iw^2

The kinetic energy is both translational and rotational

I = 2/5mr^2 and w = v/r, thus

KE rotational = .5Iw^2 = .5(2/5)(mr^2)v^2/r^2 which simplifies to .2mv^2

mgh = .5mv^2 + .2mv^2 (mass cancels)

gh = .7v^2

v = sqrt(1.43gh)

v = sqrt (1.43)(9.8)(d/sin(angle)

v = 3.74sqrt(d sin angle)

Part B)
Apply the formula above

v = 3.74 sqrt(2 sin 10)

v = 2.21 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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