A solid sphere of uniform density starts from rest and rolls without slipping a

Question

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 4.5 m down a theta = 30 degree incline. The sphere has a mass M = 3.5 kg and a radius R = 0.28 m. Of the total kinetic energy of the sphere, what fraction is translational? KE_tran/KE_total = What is the translational kinetic energy of the sphere when it reaches the bottom of the incline? KE_tran = What is the translational speed of the sphere as it reaches the bottom of the ramp? v = m/s Now let's change the problem a little. Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline? KE_tran = J

Explanation / Answer

1) KE_trans = 0.5*m*v^2

KE_total = KE_trans+KE_rot

KE_total = 0.5*m*v^2 + (0.5*I*w^2)

I is the moment of inertia of the sphere = (2/5)*m*R^2

w is the angular speed = v/R

KE_total = (0.5*m*v^2)+(0.5*(2/5)*m*R^2*(v/R)^2)

KE_total = (0.5*m*v^2)+((1/5)*m*v^2)

KE_total = 0.7*m*v^2

KE_trans/KE_total = (0.5*m*v^2)/(0.7*m*v^2) = 5/7

2) Using law of conservation of energy

energy at the top of the incline = energy at the bottom of the incline

m*g*h = KE_total

m*g*d*sin(theta) = KE_trans/(5/7)

KE_trans = (5/7)*m*g*d*sin(theta)

KE_trans = (5/7)3.5*9.8*4.5*sin(30) = 55.125 J

3) KE_trans = 0.5*m*v^2 = 55.125

0.5*3.5*v^2 = 55.125

v = 5.62 m/sec

4) if there is no friction,then there is no rotational kinetic energy

so all the energy at the bottom of the incline is KE_trans

so KE_trans = m*g*h = 3.5*9.8*4.5*sin(30) = 77.175 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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