In the post dated 20^{th} January 2008, equations to be remembered in circular motion and rotation were discussed. Subsequently some multiple choice practice questions on circular motion and rotation were discussed in the posts dated 24^{th} January 2008, 26^{th} January 2008 and 9^{th} May 2009. Free response practice questions in this section were discussed in the posts dated 23^{rd} January 2008 and March 7^{th} 2009. You can access all posts on rotational motion on this site by clicking on the

Today we will discuss a few more multiple choice questions in this section:

(1) A solid sphere of radius *r* is released (from rest) from the top inner edge (position P in fig.) of a hemispherical bowl. The sphere and the bowl have smooth surfaces. What will be the angular velocity of the sphere about the centre O of the hemispherical bowl when the sphere reaches the bottom B of the bowl?

(a) [2*g*(*R* – *r*)]^{1/2}

(b) [2*g/*(*R* – *r*)]^{1/2}

(c) [10*g/ *7(*R* – *r*)]^{1/2}

(d) [10*g/ *7(*R* – *r*)]

(e) [2*g/*5(*R* – *r*)]^{1/2}

The important thing you need to remember is that there cannot be any *rolling* in the absence of friction. The solid sphere will simply *slide* along the inner surface of the hemispherical bowl. The problem is therefore simpler than some of you might have imagined.

The centre of gravity of the sphere has come down through a distance *R* – *r* on reaching the bottom B of the hemispherical bowl. Consequently, the loss in the gravitational potential energy of the sphere is *mg*(*R* – *r*) where *m* is its mass. The sphere gains an equal amount of kinetic energy so that we have

½ *mv*^{2 }**= ***mg*(*R* – *r*) where *v* is the velocity of the sphere at the bottom B of the bowl. This gives *v* = [2*g*(*R* – *r*)]^{1/2}.

[Normally you will remember the speed *v* = √(2*gh*) in the case of a body falling freely from a height *h* (= *R* – *r* here) and you can skip the above steps while working out multiple choice questions].

Angular velocity *ω* of the sphere about the centre O of the hemispherical bowl is given by^{}

*ω = v/*(*R* – *r*) = [2*g*(*R* – *r*)]^{1/2}/(*R* – *r*) = **[2 g/(R **

**–**

*r*)]^{1/2}(2) If the surfaces of the sphere and the bowl in the above question are rough and the sphere rolls down without slipping, what will be the angular velocity of the sphere about the centre O of the hemispherical bowl when the sphere reaches the bottom B of the bowl?

(a) [2*g*(*R* – *r*)]^{1/2}

(b) [7*g/ *5(*R* – *r*)]^{1/2}

(c) [10*g/ *7(*R* – *r*)]^{1/2}

(d) [10*g/ *7(*R* – *r*)]

(e) [2*g/*5(*R* – *r*)]^{1/2}

The loss of potential energy of the sphere is *mg*(*R* – *r*) as in the above question. But the kinetic energy in this case has two parts: translational K. E. and rotational K. E. Therefore, we have

½ *mv*^{2 }+ ½ *I** ω*_{s}^{2}* ***= ***mg*(*R* – *r*) where *I* is the moment of inertia of the solid sphere * *[*I = *(2/5)*mr*^{2}] and *ω*_{s} is the spin angular velocity of the sphere (about its own axis). Since* ω*_{s} = *v/r* the above equation becomes

½ *mv*^{2 }+ ½ ×(2/5)*mr*^{2}×(*v/r*)* *^{2}* ***= ***mg*(*R* – *r*)

This gives 7*v*^{2}/10 = *g*(*R* – *r*) so that *v = *[10*g*(*R* – *r*)/ 7]^{1/2}

The angular velocity of the sphere about the centre O of the hemispherical bowl is given by

*ω = v/*(*R* – *r*) *= *[10*g*(*R* – *r*)/ 7]^{1/2}*/*(*R* – *r*) = **[10 g/ 7(R **

**–**

*r*)]^{1/2}(3) A circular disc of mass *M* and radius* R* is at rest at the top of an incline of height *H* (Fig.). On releasing, the disc rolls down the incline without slipping. What is the angular momentum of the disc about its centre of mass when it reaches the bottom of the incline?

(a) 2*MR* √(*gH*/3)^{}

(b) (*M/R*) √(*gH*/3)^{}

(c) √(2*MgH*/3)

(d) *M *√(*gH*/3R)

(e) *MR* √(*gH*/3)^{}

You have to first find out the spin angular velocity *ω* using appropriate expression for the moment of inertia *I *of the disc (*I = *½ *MR*^{2}). Additionally, you are required to calculate the angular momentum *L = Iω*.

The linear velocity *v *of the disc at the bottom of the incline is given by

½ *Mv*^{2 }+ ½ *I** ω*^{2}* ***= ***MgH*

But *v* = *ωR* so that

½ *M**ω*^{2}*R*^{2 }+ ½ ×½ *MR*^{2}*ω*^{2}* ***= ***MgH*

Or, ¾ *ω*^{2}*R*^{2} = *gH* from which *ω =*√(4*gH/*3*R ^{2}*)

Angular momentum *L *of the disc about its centre of mass is given by

*L = **Iω = *½ *MR*^{2}×√(4*gH/*3*R ^{2}*) =

*MR*√(

*gH/*3)

Good to know, thank you for sharing such a usefull tips!!

ReplyDelete