Problem 1 Part A An 6.50-cm-ciameter, 300 g solid sphere is released from rest a

Question

Problem 1 Part A An 6.50-cm-ciameter, 300 g solid sphere is released from rest at the top of a 2.00m rg, 150°?cire it rols, without slipping, to the botiom. What is the sphere's angular velocity at the bottom of the incline? Express your answer with the appropriate units You may want to review (Pages 315-317 828? Correct Here we leam how to calculate the angular velocity of the rolling object knowing its shape, dameter, angle of inclination and length of path Part B Whait fraction of its kinetic energy is rotational? 70.94 Submit X Incorrect Try Again; 2 attempts remaining

Explanation / Answer

First part of the problem you have already solved.

From the first part, ? = 82.8 rad/s

Now, the total kinetic energy = potential energy = m*g*h = m*g*L*sin15

                                              = 0.300*9.81*2.0*0.259 = 0.762 J

Rotational KE = ½ * I * ?^2
For a solid sphere, I = 2/5 * m * r^2
Rotational KE = ½ * 2/5 * m * r^2 * ?^2 = 0.2 * m * r^2 * ?^2  = 0.2*0.300*0.0325^2*82.8^2

                                                                                                  = 0.434 J

Therefore, the fraction = 0.434 / 0.762 = 0.5695 = 0.57

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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