A uniform disk of mass M and radius R is free to rotate in the xy-plane about a

Question

A uniform disk of mass M and radius R is free to rotate in the xy-plane about a horizontal smooth axis through a pivot P on its circumfcrence as shown in the figure at right. What is the magnitude of the angular acceleration of the disk about the axis of rotation (axis through P parallel to z-axis) when line OP makes an angle 0 with the vertical (y-axis)? A uniform solid sphere of of mass 6 kg rolls without slipping on a horizontal surface. At a certain instant, its center of mass has a speed of 20 m/s. The total kinetic energy of the sphere at that instant is:

Explanation / Answer

6) M.I. about center of mass of disc= MR2/2

using parallel axcis thm: M.I. about P is = Icm + MR2 = 3MR2/2

torqua acting about P due to its weight = MgR sin(theta)

==> angulat acceleration = torque about P / M.I. about P = 2g sin (theta)/(3R)

7) since the sphere is rolling ==> velocity of center of mass Vcm = angular velocity x Radius

==> angular velocity w= Vcm/R

total K.E. = translational KE + rotational KE = MVcm2/2 + Icm w2/2

Icm for solid sphere = 2MR2/5 & substituing w= Vcm/R

==> total KE = 1/2 (MVcm2 + 2MVcm2/5) = 7MVcm2/10 = 7 x 6 x 400 / 10 =1680 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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