A hollow sphere of radius 0.180 m, with rotational inertia I = 0.0980 kg·m2 abou

Question

A hollow sphere of radius 0.180 m, with rotational inertia I = 0.0980 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 16.8° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 33.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 0.550 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Explanation / Answer

There are two component in the kinetic energy of the sphere. rotational and linear. K_rot=0.5*I*w^2. K_li=0.5*mv^2=0.5*m*r^2*w^2. moment of inertia of a hollow sphere. I=2mr^2/3. ----- m = 3I/2r^2=4.5(kg) so K_rot/K_li=2/3 so K_rot/K_total=2/5 a) so K_rot=33*2/5=13.2(J) b) K_rot=0.5*I*w^2 so w=16.4(rad/s). so v=w*r=16.4*0.18=3(m/s) c)conservation of energy K_total'=K_total-4.5*g*0.55=8.75(J). -- K_li=3K_total'/5=5.25(J) K_li=0.5*m*v^2 so v=1.5(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.