A uniform solid disk of radius R and mass M is free to rotate on a frictionless

Question

A uniform solid disk of radius R and mass M is free to rotate on a frictionless
pivot through a point on its rim. If it starts from rest with its center at the same
height as the pivot:
(a) What is its angular acceleration both at the start and when its center is directly
below the pivot ?
(b) What is its angular velocity both at the start and when its center is directly below
the pivot ?
(c) What are the components of the reaction force of the pivot on the disk both at the
start and when its center is directly below the pivot

Explanation / Answer

a)angular acceleration. a=torque/moment of inertia at start. torque = M*R*g. moment of inertia I=MR^2/2+MR^2=3MR^2/2 so a=MRg/(3MR^2/2)=2g/3R. below the pivot. a=0. because T=0. b) start w=0. below the pivot conservation of energy. Mg*R=I*w^2/2. so w=sqrt(2MgR/I)=sqrt(2MgR/3MR^2/2)=sqrt(4g/3R) c) when it is at rest, F=Mg-2g*M/3=Mg/3. directly below. F=Mg+Mw^2*R=Mg+M*4g/3=7Mg/3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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