A solid, uniform sphere of mass 2.0 kg and radius 1.7 m starts from rest and rol

Question

A solid, uniform sphere of mass 2.0 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of height 5 m. What is the angular velocity of the sphere at the bottom of the inclined plane? A bowling ball rolls without slipping, first along a level track, then up a ramp onto another level section of the track, gaining 0.340 m in altitude. If its translational speed along the lower track level was 3.15 m/s, find its translational speed at the top. A basketball of mass m and radius r rolls without slipping up a hill. The angle between the hill and the horizontal is theta . The initial speed of the basketball is v. The magnitude of the acceleration due to gravity is g. Expressed in terms of m, r, theta , v and/or g, what vertical height, h, does the ball gain before it comes to rest (momentarily)? Assume that the ball is a rigid body, and that friction and air resistance can be ignored.

Explanation / Answer

a)here I=2/5 *mr^2=2.312 kgm2.... energy balance for no slip v=rw 0.5*v^2+0.5*Iw^2/m=gh=9.8*5 angular vel,w=4.9215 rad/s b)from energy blanace 0.5*v^2=0.5*3.15^2+9.8*0.34 v=4.07 m/s c)from energy balance at h 0.5*v^2+0.5*Iw^2/m=gh ...for basket ball I=2/3 *mr^2 for no slip v=rw 0.5v^2+0.5*2/3 *v^2 =gh h=5/6 *v^2/g=0.085 *v^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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