AP Physics C- Additional Multiple Choice Practice Questions on Rotational Motion

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 label ‘rotation’ below this post.

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) [2g(R – r)]^1/2

(b) [2g/(R – r)]^1/2

(c) [10g/ 7(R – r)]^1/2

(d) [10g/ 7(R – r)]

(e) [2g/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 mv² = mg(R – r) where v is the velocity of the sphere at the bottom B of the bowl. This gives v = [2g(R – r)]^1/2.

[Normally you will remember the speed v = √(2gh) 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) = [2g(R – r)]^1/2/(R – r) = [2g/(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) [2g(R – r)]^1/2

(b) [7g/ 5(R – r)]^1/2

(c) [10g/ 7(R – r)]^1/2

(d) [10g/ 7(R – r)]

(e) [2g/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 mv² + ½ I ω_s² = mg(R – r) where I is the moment of inertia of the solid sphere [I = (2/5)mr²] 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 mv² + ½ ×(2/5)mr²×(v/r) ² = mg(R – r)

This gives 7v²/10 = g(R – r) so that v = [10g(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) = [10g(R – r)/ 7]^1/2/(R – r) = [10g/ 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) 2MR √(gH/3)

(b) (M/R) √(gH/3)

(c) √(2MgH/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 MR²). 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 Mv² + ½ I ω² = MgH

But v = ωR so that

½ Mω²R² + ½ ×½ MR²ω² = MgH

Or, ¾ ω²R² = gH from which ω =√(4gH/3R²)

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

L = Iω = ½ MR²×√(4gH/3R²) = MR√(gH/3)

Answer to AP Physics C Free Response Practice Question on Electric Field & Potential

In the post dated 2^nd August 2009, the following free-response question for practice was given to you:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

As promised, I give below a model answer for the above question:

(a) The electric field at the centre of the ring is zero.

This follows from the definition of the electric field: Electric field at any point is the force per unit positive test charge placed at the point. The test charge placed at the centre of the ring will be repelled equally by the uniformly distributed positive charges on the conducting ring so that the net force on the test charge will be zero.

[Note that the charge Q placed at the position A on the conducting ring will be uniformly distributed immediately throughout the ring].

(b):

(i) The charges on the ring are at the same distance R from the centre of the ring and hence the potential at the centre of the ring is Q/4πε₀R.

(ii) If the charge placed on the ring is negative, the potential will be negative. The potential in this case will be – Q/4πε₀R.

(c) Consider a small quantity (dQ) of charge at a point such as A on the ring. The electric potential dV at P due to this elemental charge is given by

dV = dQ/4πε₀r.

The total potential (V) at P due to all the charges on the ring is given by

V = ∫dV = ∫dQ/4πε₀r = Q/4πε₀r (since all charge elements are at the same distance r from the point P).

(d) The electric field at P due to an element dQ of charge at a point such as A is directed along AP. This field can be resolved into two components: one component along the axis of the ring (axial component) and the other component perpendicular to the axis of the ring (normal component). For every elemental charge at any given point on the ring, there is an equal elemental charge situated diametrically opposite to it. The diametrically opposite element will produce an equal axial component of field in the same direction. But the normal component due to the diametrically opposite elemental charge will be equal in magnitude but opposite in direction. Therefore, all axial components get added where as all normal components get canceled.

The electric field due to the charged conducting ring is therefore directed along the axis of the ring.

The electric field at the point P is given by

E = –grad V.

Since we know that the electric field is along the axis of the ring which is along the x-axis as given in the question, we have

E = –∂V/∂x = –∂/_∂x (Q/4πε₀r) = –∂/_∂x [Q/4πε₀√(R²+x²)]

Or, E = (1/4πε₀) [Qx/(R²+x²)^3/2]

Since the charge on the ring is positive, the field E is along the positive x-direction.

When the electric field is maximum, dE/dx = 0

Therefore, (1/4πε₀)[{(–3/2)(R²+x²)^–5/2 ×2x ×x} + (R²+x²)^–3/2] = 0

Or, 3x² (R²+x²)^–1 = 1

Or, 3x² = R²+x² from which x = R/√2.

(e) The nature of variation of the electric field along the axis of the ring is shown in the adjoining figure in which O represents the centre of the ring. The field has maximum magnitude at points distant R/√2 from the centre and these are located symmetrically on either side of the centre. Positive direction of the field is the positive x-direction even though it is represented by positive y-plot in the figure. The negative direction of the field is the negative x-direction and is represented by negative y-plot in the figure.

You can access all posts related to electrostatics (including multiple choice practice questions) on this site by clicking on the label ‘electrostatics’ below this post.

Electrostatics- AP Physics C Free Response Practice Question on Electric Field & Potential

Essential points to be remembered in respect of electric field and potential were discussed on this blog on 16^th June 2008. You can access that post by clicking here.

All posts related to electrostatics can be accessed by clicking on the label ‘electrostatics’ below this post.

Today I’ll give you a free response practice question involving electric field and potential:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

This question carries 15 points and you have 15 minutes for answering it. Try to answer this question. I’ll be back soon with a model answer for your benefit.

AP Physics B & C – Multiple Choice Questions (MCQ) for Practice on Gravitation

Equations to be remembered in respect of the section on gravitation were given in the post dated 9^th May 2008. A few multiple choice practice questions were discussed in the post dated 12^th May 2008 followed by a free response practice question on gravitation in the post dated 15^th May 2008. You can access all posts on gravitation on this site by clicking on the label ‘gravitation’ below this post.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are meant for AP Physics B as well as C:

(1) The speed of an artificial satellite moving in an orbit of radius r around the earth is increased by 41.4%. The satellite will

(a) continue to move in the same orbit

(b) move in an orbit of radius 1.414 r

(c) move in an orbit of radius 0.414 r

(d) move in an orbit of radius 0.586 r

(e) escape into the outer space

The orbital speed v of any satellite is given by

v =√(gr) where g is the acceleration due to gravity at the orbit of radius r.

[You will get this by equating the centripetal force to the gravitational pull:

mv²/r = mg]

The escape velocity (v_escape) of a body at a point distant r from the centre of the earth where the acceleration due to gravity is equal to g is given by

v_escape =√(2gr)

Thus the escape velocity is √2 times the orbital speed.

The speed v of the satellite becomes 1.414 times the initial value on increasing the speed by 41.4%. Now, 1.414 v = √2 v. Therefore the new speed is the escape speed so that the satellite will escape into the outer space.

(2) A launch vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. What is the minimum energy to be spent by the launch vehicle on the satellite so that the satellite will move in a circular orbit of radius 7R? (Gravitational constant = G)

(a) GMm /7R

(b) 6GMm /7R

(c) 6GMm /14R

(d) 13GMm /14R

(e) GMm /14R

The satellite of mass m is initially at rest and therefore its initial energy (E_i) is its gravitational potential energy given by

E_i = – GMm/R

When the satellite is in its orbit of radius 7R, its final energy (E_f) is given by

E_f = – GMm/(2×7R) = – GMm/(14R)

[Note that the total energy of a satellite in an orbit of radius r is – GMm/2r]

The minimum energy required for placing the satellite in its orbit is E_f – E_i given by

E_f – E_i = – [GMm/(14R)] – [– GMm/R]

= (GMm/R) – [GMm/(14R)]

= 13GMm /14R

(3) A small body is projected vertically up with a speed equal to half the escape speed from the earth’s surface. If the radius of the earth is R, what is the maximum height (h) reached by the body? (Neglect air resistance).

(a) R/3

(b) R/2

(c) R

(d) 2R

(e) 5R

When the body projected vertically reaches the maximum height, its speed will be zero since its initial kinetic energy us used up in increasing its gravitational potential energy. Therefore we have

½ m (v_e/2)² = [–GMm /(R+h)] – [–GMm /R] where m is the mass of the body, M is the mass of the earth, v_e is the escape speed from the earth’s surface and G is the gravitational constant.

[The first term on the right hand side of the above equation is the gravitational potential energy of the body at height h. The second term is the gravitational potential energy of the body on the surface of the earth].

Since v_e = √(2GM/R), the above equation becomes

GM /4R = (GM /R) –[GM /(R+h)]

Or, GM /4R = GMh /[R(R+h)

Therefore, ¼ = h/(R+h) from which h = R/3

(4) The orbit of a planet moving around the sun is elliptical with the sun S at one focus of the ellipse. If the orbital speeds of the planet while at points A and B (Fig.) are V₁ and V₂ and the distances SA and SB are d₁ and d₂ respectively, V₁/ V₂ is equal to

(a) √(d₁ /d₂)

(b) √(d₂ /d₁)

(c) d₁ /d₂

(d) d₂ /d₁

(e) (d₂ + d₁) /(d₂ – d₁)

To obtain the answer you may apply Kepler’s law of areas (in respect of plnetary motion), which says that the straight line joining the sun to the planet sweeps equal areas in equal intervals of time. The adjoining figure shows path lengths CD and EF traced by the planet in equal intervals (say, ∆t) and hence the areas CADS and EBFS must be equal. If ∆t is sufficiently small, CD and EF will become straight line segments and the areas will become triangular. Equating the areas of these triangles, we have

½ ×CD×AS = ½ ×EF×BS

Or, ½ × V₁ ∆t × d₁ = ½ × V₂ ∆t × d₂

Therefore, V₁/ V₂ = d₂ /d₁

[You can easily get the answer from angular momentum conservation:

m V₁ d₁ = m V₂ d₂

where m is the mass of the planet. This gives V₁/ V₂ = d₂ /d₁]

(5) We know that the gravitational force between two point masses is inversely proportional to the square of the distance between them. If the gravitational force between two point masses were (let us imagine) inversely proportional to the n^th power of the distance between them, what would be the relation between the orbital period (T) of a planet around the sun and the mean distance (r) of the planet from the sun?

(a) T² α r²ⁿ

(b) T² α rⁿ

(c) T² α r²ⁿ⁺¹

(d) T² α r⁽ⁿ⁺¹⁾

(e) T² α r^{(n – 1)}

For simplicity, let us assume that the orbit is circular. (Circular orbit is only a special case of the general elliptical orbit and the relation between T and r will be unchanged). The centripetal force required for the circular motion is supplied by the gravitational pull so that we have

mrω² = GMm/rⁿ where m is the mass of the planet, ω is its orbital angular velocity, M is the mass of the sun and G is the gravitational constant.

The above equation gives ω² = GM/r⁽ⁿ⁺¹⁾

Since ω = T/2π, we have

(T/2π)² = GM/r⁽ⁿ⁺¹⁾

Therefore, T² α r⁽ⁿ⁺¹⁾

You will find some useful multiple choice questions on gravitation (with solution) at physicsplus.