Circular Motion Quiz: 50 Physics Questions with Detailed Solutions

A particle moves in a circle of radius 4 m with a constant speed of 8 m/s. What is its centripetal acceleration?

1. 16 m/s²
2. 18 m/s²
3. 32 m/s²
4. 64 m/s²

Explanation

The centripetal acceleration is

\(a_c=\dfrac{v^2}{r}\)

\(a_c=\dfrac{8^2}{4}=\dfrac{64}{4}=16\;m/s^2\)

Therefore, the correct answer is 16 m/s².

A 2 kg object moves in a horizontal circle of radius 5 m at a speed of 10 m/s. What is the required centripetal force?

1. 20 N
2. 40 N
3. 50 N
4. 60 N

Explanation

The centripetal force is

\(F_c=\dfrac{mv^2}{r}\)

\(F_c=\dfrac{2\times10^2}{5}=\dfrac{200}{5}=40N\)

Therefore, the required force is 40 N.

A wheel rotates at 5 revolutions per second. What is its angular velocity?

1. 5π rad/s
2. 15π rad/s
3. 20π rad/s
4. 10π rad/s

Explanation

Angular velocity is

\(\omega=2\pi f\)

\(\omega=2\pi(5)=10\pi\;rad/s\)

Therefore, the correct answer is 10π rad/s.

A car travels around a circular track of radius 50 m at a speed of 20 m/s. What is its centripetal acceleration?

1. 4 m/s²
2. 8 m/s²
3. 6 m/s²
4. 10 m/s²

Explanation

Using

\(a_c=\dfrac{v^2}{r}\)

\(a_c=\dfrac{20^2}{50}=\dfrac{400}{50}=8\;m/s^2\)

Therefore, the answer is 8 m/s².

A stone tied to a string moves in a circle. If the string suddenly breaks, the stone will move in which direction?

1. Along the tangent at the point of release
2. Toward the center
3. Away from the center
4. In another circular path

Explanation

The centripetal force disappears when the string breaks. Due to inertia, the stone continues along the direction of its instantaneous velocity, which is tangent to the circle.

A satellite orbits Earth in a circular orbit of radius \(7\times10^6\) m. If \(g=8.14\;m/s^2\) at that altitude, what is its orbital speed?

1. 5000 m/s
2. 6500 m/s
3. 7550 m/s
4. 9000 m/s

Explanation

For circular orbit,

\(\dfrac{v^2}{r}=g\)

\(v=\sqrt{gr}\)

\(v=\sqrt{8.14\times7\times10^6}\)

\(v\approx7550\;m/s\)

Therefore, the correct answer is approximately 7550 m/s.

A car moves on a level circular road of radius 100 m. If the coefficient of static friction is 0.25 and \(g=10\;m/s^2\), what is the maximum speed without skidding?

1. 10 m/s
2. 12 m/s
3. 14 m/s
4. 15.8 m/s

Explanation

Maximum friction provides the centripetal force.

\(\mu mg=\dfrac{mv^2}{r}\)

\(v=\sqrt{\mu gr}\)

\(v=\sqrt{0.25\times10\times100}\)

\(v=\sqrt{250}\)

\(v\approx15.8\;m/s\)

A particle completes one revolution in 4 seconds. What is its angular velocity?

1. \(\pi/4\) rad/s
2. \(\pi\) rad/s
3. \(2\pi\) rad/s
4. \(\pi/2\) rad/s

Explanation

Angular velocity is

\(\omega=\dfrac{2\pi}{T}\)

\(\omega=\dfrac{2\pi}{4}\)

\(\omega=\dfrac{\pi}{2}\;rad/s\)

A conical pendulum has a string length of 2 m and makes an angle of 60° with the vertical. What is the radius of the circular path?

1. 0.5 m
2. 1.0 m
3. 1.73 m
4. 2.0 m

Explanation

The radius is

\(r=L\sin\theta\)

\(r=2\sin60^\circ\)

\(r=2\times0.866\)

\(r=1.732\;m\)

Therefore, the answer is 1.73 m.

For a particle moving with constant angular velocity \(\omega\), which statement is correct?

1. Acceleration is zero
2. Velocity is constant
3. Speed is constant but velocity changes continuously
4. Both speed and velocity remain constant

Explanation

In uniform circular motion, the magnitude of velocity remains constant. However, its direction changes continuously. Therefore, velocity changes and centripetal acceleration exists.

A particle moves in a circle of radius 10 m with a speed of 20 m/s. What is the centripetal acceleration?

1. 10 m/s²
2. 20 m/s²
3. 40 m/s²
4. 80 m/s²

Explanation

Using

\(a_c=\dfrac{v^2}{r}\)

\(a_c=\dfrac{20^2}{10}=\dfrac{400}{10}=40\;m/s^2\)

Therefore, the answer is 40 m/s².

A 0.5 kg ball is tied to a string and rotated in a horizontal circle of radius 2 m at 4 m/s. What is the centripetal force?

1. 2 N
2. 4 N
3. 4 N
4. 8 N

Explanation

\(F_c=\dfrac{mv^2}{r}\)

\(F_c=\dfrac{0.5\times4^2}{2}\)

\(F_c=\dfrac{0.5\times16}{2}=4N\)

Therefore, the centripetal force is 4 N.

A wheel rotates at 120 rpm. What is its angular velocity?

1. \(2\pi\) rad/s
2. \(3\pi\) rad/s
3. \(4\pi\) rad/s
4. \(8\pi\) rad/s

Explanation

120 rpm = 2 rev/s

\(\omega=2\pi f\)

\(\omega=2\pi\times2=4\pi\;rad/s\)

A particle moves in a circle of radius 5 m with angular velocity 6 rad/s. What is its linear speed?

1. 20 m/s
2. 25 m/s
3. 30 m/s
4. 35 m/s

Explanation

\(v=\omega r\)

\(v=6\times5=30\;m/s\)

Therefore, the correct answer is 30 m/s.

A car rounds a curve of radius 80 m at 16 m/s. What centripetal acceleration does it experience?

1. 2.4 m/s²
2. 2.8 m/s²
3. 3.0 m/s²
4. 3.2 m/s²

Explanation

\(a_c=\dfrac{v^2}{r}\)

\(a_c=\dfrac{16^2}{80}\)

\(a_c=\dfrac{256}{80}=3.2\;m/s^2\)

A stone moves in a vertical circle. At the highest point, the minimum speed required to keep the string taut is:

1. \(\sqrt{rg/2}\)
2. \(\sqrt{2rg}\)
3. \(\sqrt{rg}\)
4. \(rg\)

Explanation

At the topmost point, minimum tension is zero.

\(\dfrac{mv^2}{r}=mg\)

Therefore,

\(v=\sqrt{rg}\)

A conical pendulum has a string length of 1.5 m and makes an angle of 30° with the vertical. What is the radius of the circular path?

1. 0.50 m
2. 0.75 m
3. 1.00 m
4. 1.50 m

Explanation

\(r=L\sin\theta\)

\(r=1.5\times\sin30^\circ\)

\(r=1.5\times0.5=0.75\;m\)

A satellite moves in a circular orbit around Earth. Which quantity remains constant during uniform circular motion?

1. Velocity
2. Centripetal acceleration direction
3. Speed
4. Force direction

Explanation

In uniform circular motion, speed remains constant. However, velocity, acceleration direction, and force direction continuously change.

A particle completes 50 revolutions in 10 seconds. What is its frequency?

1. 2 Hz
2. 5 Hz
3. 10 Hz
4. 25 Hz

Explanation

Frequency is

\(f=\dfrac{N}{t}\)

\(f=\dfrac{50}{10}=5\;Hz\)

A road is banked at an angle of 45°. Neglecting friction and taking \(g=10\;m/s^2\), what is the speed of a car moving on a curve of radius 50 m?

1. 15 m/s
2. 18 m/s
3. 22.4 m/s
4. 30 m/s

Explanation

For a frictionless banked road,

\(\tan\theta=\dfrac{v^2}{rg}\)

Since

\(\tan45^\circ=1\)

\(v=\sqrt{rg}\)

\(v=\sqrt{50\times10}\)

\(v=\sqrt{500}\)

\(v\approx22.4\;m/s\)

Therefore, the correct answer is 22.4 m/s.

A particle moves in a horizontal circle of radius 2 m such that its speed varies with time as \(v(t)=3t^2\) m/s. What is the centripetal acceleration at \(t=2\) s?

1. 18 m/s²
2. 36 m/s²
3. 72 m/s²
4. 144 m/s²

Explanation

At \(t=2\) s,

\(v=3(2)^2=12\;m/s\)

Centripetal acceleration:

\(a_c=\dfrac{v^2}{r}=\dfrac{12^2}{2}=72\;m/s^2\)

A particle moves along a circular path of radius 4 m. Its angular position is given by \(\theta=t^3\), where \(\theta\) is in radians. What is the tangential acceleration at \(t=2\) s?

1. 12 m/s²
2. 24 m/s²
3. 48 m/s²
4. 96 m/s²

Explanation

\(\omega=\dfrac{d\theta}{dt}=3t^2\)

\(\alpha=\dfrac{d\omega}{dt}=6t\)

At \(t=2\),

\(\alpha=12\;rad/s^2\)

Tangential acceleration:

\(a_t=r\alpha=4\times12=48\;m/s^2\)

A particle moves in a circle of radius 5 m with constant angular acceleration \(2\;rad/s^2\). If its initial angular velocity is \(4\;rad/s\), what is its speed after 3 s?

1. 30 m/s
2. 40 m/s
3. 50 m/s
4. 60 m/s

Explanation

\(\omega=\omega_0+\alpha t\)

\(\omega=4+2\times3=10\;rad/s\)

\(v=r\omega=5\times10=50\;m/s\)

A bead slides frictionlessly on a circular wire of radius R placed vertically. At the highest point, what is the minimum speed required so that the bead does not lose contact?

1. \(\sqrt{\dfrac{gR}{2}}\)
2. \(\sqrt{2gR}\)
3. \(\sqrt{gR}\)
4. \(gR\)

Explanation

At the highest point, minimum contact occurs when the normal force becomes zero.

\(\dfrac{mv^2}{R}=mg\)

Thus,

\(v=\sqrt{gR}\)

A conical pendulum has string length 1 m and angular speed \(\omega=5\;rad/s\). What angle does the string make with the vertical? (\(g=10\;m/s^2\))

1. \(cos^{-1}\left(\frac{2}{3}\right)\)
2. \(cos^{-1}\left(\frac{1}{4}\right)\)
3. \(cos^{-1}\left(\frac{2}{5}\right)\)
4. \(cos^{-1}\left(\frac{1}{5}\right)\)

Explanation

For a conical pendulum,

\(\omega^2=\dfrac{g}{L\cos\theta}\)

\(25=\dfrac{10}{\cos\theta}\)

\(\cos\theta=\dfrac{2}{5}\)

\(\theta = cos^{-1}\left(\dfrac{2}{5}\right)\)

A particle moves in a circle of radius 3 m. Its angular velocity varies as \(\omega=2t+1\). What is the magnitude of total acceleration at \(t=2\) s?

1. 45 m/s²
2. 60 m/s²
3. 75 m/s²
4. 90 m/s²

Explanation

At \(t=2\),

\(\omega=5\;rad/s\)

\(\alpha=2\;rad/s^2\)

Tangential acceleration:

\(a_t=r\alpha=3\times2=6\)

Centripetal acceleration:

\(a_c=r\omega^2=3\times25=75\)

Total acceleration:

\(\sqrt{75^2+6^2}\approx75.2\)

Closest answer: 75 m/s²

A satellite orbits Earth at a height equal to Earth’s radius. If orbital speed near the surface is \(v_0\), what is the orbital speed at this height?

1. \(v_0\)
2. \(\dfrac{v_0}{\sqrt2}\)
3. \(\dfrac{v_0}{\sqrt2}\)
4. \(\dfrac{v_0}{2}\)

Explanation

Orbital speed:

\(v_0=\sqrt{\dfrac{GM}{r}}\)

At height \(R\),

Total distance from center \(=2R\)

\(v=\sqrt{\dfrac{GM}{2R}}=\dfrac{v_0}{\sqrt2}\)

A particle is projected horizontally from the top of a smooth hemisphere of radius R. At what speed will it lose contact immediately?

1. 0
2. \(\sqrt{\dfrac{gR}{2}}\)
3. \(\sqrt{gR}\)
4. \(\sqrt{2gR}\)

Explanation

Loss of contact occurs when the normal reaction becomes zero.

\(mg=\dfrac{mv^2}{R}\)

Therefore,

\(v=\sqrt{gR}\)

A cyclist rides inside a vertical cylindrical wall of radius 8 m. If the coefficient of friction is 0.5, what is the minimum speed required to avoid sliding down? (\(g=10\;m/s^2\))

1. 8 m/s
2. 12.6 m/s
3. 16 m/s
4. 20 m/s

Explanation

Maximum friction:

\(f=\mu N\)

For equilibrium,

\(\mu N=mg\)

Since

\(N=\dfrac{mv^2}{r}\)

\(\mu\dfrac{mv^2}{r}=mg\)

\(v=\sqrt{\dfrac{rg}{\mu}}\)

\(v=\sqrt{\dfrac{8\times10}{0.5}}\)

\(v=\sqrt{160}\)

\(v\approx12.6\;m/s\)

A particle moves under the force \(\vec F=-m\omega^2\vec r\). Which type of trajectory is possible if the motion is confined to one dimension?

1. Simple harmonic motion
2. Uniform circular motion
3. Parabolic motion
4. Hyperbolic motion

Explanation

The force is directly proportional to displacement and opposite in direction.

\(\vec F=-m\omega^2\vec r\)

This is the standard restoring-force equation of SHM.

A particle moves in a vertical circle of radius 2 m. What is the minimum speed at the lowest point so that it can complete the entire circle? (Take \(g=10\;m/s^2\))

1. 5 m/s
2. \(4\sqrt{5}\) m/s
3. 10 m/s
4. \(2\sqrt{10}\) m/s

Explanation

For just completing the circle:

At the top,

\(v_{top}=\sqrt{gr}\)

Using conservation of energy between bottom and top:

\[
\frac12 mv_b^2=\frac12 mv_t^2+mg(2r)
\]

\[
v_b^2=gr+4gr=5gr
\]

\[
v_b=\sqrt{5gr}
\]

\[
v_b=\sqrt{5\times10\times2}=10\;m/s
\]

Therefore,

\(v_b=2\sqrt{25}=10\;m/s\)

A particle moves in a horizontal circle of radius 5 m. Its angular position is given by \(\theta=2t^3-3t\). What is the angular acceleration at \(t=2\) s?

1. 12 rad/s²
2. 18 rad/s²
3. 20 rad/s²
4. 24 rad/s²

Explanation

\[
\omega=\dfrac{d\theta}{dt}=6t^2-3
\]

\[
\alpha=\dfrac{d\omega}{dt}=12t
\]

At \(t=2\),

\[
\alpha=24\;rad/s^2
\]

A particle moves in a circle of radius 4 m with angular velocity \(\omega=t^2+2\). Find the magnitude of total acceleration at \(t=2\) s.

1. 70 m/s²
2. 145 m/s²
3. 160
4. 200 m/s²

Explanation

\[
\omega=2^2+2=6\;rad/s
\]

Centripetal acceleration:

\[
a_c=r\omega^2=4(36)=144
\]

Angular acceleration:

\[
\alpha=2t=4
\]

Tangential acceleration:

\[
a_t=r\alpha=16
\]

Total acceleration:

\[
a=\sqrt{144^2+16^2}
\]

\[
a\approx145\;m/s^2
\]

A small block is placed on a smooth hemisphere of radius R. At what angle \(\theta\) from the vertical does it lose contact with the surface?

1. 30°
2. 45°
3. \(\cos\theta=\frac23\)
4. \(\cos\theta=\frac12\)

Explanation

At separation:

\[
N=0
\]

Radial equation:

\[
mg\cos\theta=\dfrac{mv^2}{R}
\]

Energy conservation:

\[
mgR(1-\cos\theta)=\frac12 mv^2
\]

Combining both equations:

\[
\cos\theta=\frac23
\]

A satellite revolves around Earth in a circular orbit of radius \(4R\), where \(R\) is Earth’s radius. If the orbital period near the surface is \(T\), what is the new period?

1. \(4T\)
2. \(6T\)
3. \(8T\)
4. \(16T\)

Explanation

Kepler’s Third Law:

\[
T^2\propto r^3
\]

\[
\dfrac{T_2}{T_1}
=
\left(\dfrac{4R}{R}\right)^{3/2}
\]

\[
=8
\]

Therefore,

\(T_2=8T\)

A bead moves in a smooth circular ring rotating with constant angular speed \(\omega\) about a vertical diameter. The bead remains at rest relative to the ring. The equilibrium angle \(\theta\) satisfies:

1. \(\cos\theta=\dfrac{\omega^2R}{g}\)
2. \(\cos\theta=\dfrac{g}{\omega^2R}\)
3. \(\sin\theta=\dfrac{g}{\omega^2R}\)
4. \(\tan\theta=\dfrac{g}{\omega^2R}\)

Explanation

Resolving effective forces along the ring:

\[
mg=\omega^2R\cos\theta
\]

Hence,

\[
\cos\theta=\dfrac{g}{\omega^2R}
\]

A particle moves in a circle with speed \(v\). If both the speed and radius become three times their original values, the centripetal acceleration becomes:

1. 3 times
2. 6 times
3. 9 times
4. 27 times

Explanation

\[
a_c=\dfrac{v^2}{r}
\]

New acceleration:

\[
a’=\dfrac{(3v)^2}{3r}
=\dfrac{9v^2}{3r}
=3a
\]

Therefore, centripetal acceleration becomes 3 times the original value.

A particle moves in a vertical circle. At a certain point the string tension is three times the weight of the particle. If the velocity at that point is horizontal, what is the centripetal acceleration?

1. \(g\)
2. \(2g\)
3. \(3g\)
4. \(4g\)

Explanation

At the side point, weight has no radial component.

\[
T=\dfrac{mv^2}{r}
\]

Given

\[
T=3mg
\]

Thus

\[
\dfrac{mv^2}{r}=3mg
\]

\[
a_c=\dfrac{v^2}{r}=3g
\]

A particle is attached to a string and rotated in a horizontal circle. If the tension is doubled while radius remains unchanged, the speed becomes:

1. \(\sqrt2\,v\)
2. \(\sqrt2\,v\)
3. \(2v\)
4. \(4v\)

Explanation

\[
T=\dfrac{mv^2}{r}
\]

If tension becomes \(2T\),

\[
2T=\dfrac{mv’^2}{r}
\]

\[
v’^2=2v^2
\]

\[
v’=\sqrt2\,v
\]

A particle moves in a circle of radius 1 m with angular velocity \(\omega=4t\). At \(t=1\) s, the angle between total acceleration and radius vector is closest to:

1. 14°
2. 26.5°
3. 45°
4. 63.5°

Explanation

At \(t=1\),

\[
\omega=4
\]

\[
a_c=r\omega^2=16
\]

\[
\alpha=4
\]

\[
a_t=r\alpha=4
\]

Angle with radius:

\[
\tan\theta=\dfrac{a_t}{a_c}
=\frac4{16}
=\frac14
\]

\[
\theta\approx14^\circ
\]

Since total acceleration is measured from the inward radial direction, the closest answer is 14°.

A small bead slides without friction on the outer surface of a fixed sphere of radius R. It is released from rest from the top. At what angle θ from the vertical does the bead lose contact with the sphere?

1. \(\cos\theta=\dfrac{1}{2}\)
2. \(\cos\theta=\dfrac{2}{3}\)
3. \(\cos\theta=\dfrac{3}{4}\)
4. \(\cos\theta=\dfrac{1}{3}\)

Explanation

Using conservation of energy,

\[
v^2=2gR(1-\cos\theta)
\]

At the point of separation,

\[
N=0
\]

Therefore,

\[
mg\cos\theta=\dfrac{mv^2}{R}
\]

Substituting the energy equation,

\[
g\cos\theta=2g(1-\cos\theta)
\]

\[
3\cos\theta=2
\]

\[
\cos\theta=\frac23
\]

A particle is attached to a string and moves in a vertical circle of radius R. What is the minimum speed at the lowest point required to complete the circle?

1. \(\sqrt{3gR}\)
2. \(2\sqrt{gR}\)
3. \(\sqrt{5gR}\)
4. \(\sqrt{6gR}\)

Explanation

For just-complete motion,

\[
v_{top}=\sqrt{gR}
\]

Using conservation of energy between bottom and top,

\[
\frac12 mv_b^2
=
\frac12 mgR+2mgR
\]

\[
v_b^2=5gR
\]

\[
v_b=\sqrt{5gR}
\]

A car moves on a banked road of angle θ without friction. Which speed allows the car to negotiate the curve of radius R without slipping?

1. \(v=\sqrt{Rg\cos\theta}\)
2. \(v=\sqrt{Rg\tan\theta}\)
3. \(v=\sqrt{\dfrac{Rg}{\tan\theta}}\)
4. \(v=\sqrt{Rg\sin\theta}\)

Explanation

Resolving forces,

\[
N\cos\theta=mg
\]

\[
N\sin\theta=\dfrac{mv^2}{R}
\]

Dividing,

\[
\tan\theta=\dfrac{v^2}{Rg}
\]

Thus,

\[
v=\sqrt{Rg\tan\theta}
\]

A satellite moves in a circular orbit of radius r around Earth. If the orbital radius becomes 9r, the orbital period becomes:

1. 9T
2. 27T
3. 3T
4. 81T

Explanation

Kepler’s third law gives

\[
T^2\propto r^3
\]

Hence,

\[
\dfrac{T’}{T}
=
\left(\dfrac{9r}{r}\right)^{3/2}
=
27
\]

A particle moves in a circle of radius R with constant angular acceleration α. Initially it is at rest. After rotating through an angle θ, its angular speed is:

1. \(\omega=\alpha\theta\)
2. \(\omega=\sqrt{2\alpha\theta}\)
3. \(\omega=2\alpha\theta\)
4. \(\omega=\sqrt{\alpha\theta}\)

Explanation

Rotational kinematics gives

\[
\omega^2=\omega_0^2+2\alpha\theta
\]

Since

\[
\omega_0=0
\]

\[
\omega=\sqrt{2\alpha\theta}
\]

A bead is placed on a smooth circular wire rotating with angular velocity ω about its vertical diameter. The bead can remain at a position other than the lowest point only if:

1. \(\omega<\sqrt{\dfrac{g}{R}}\)
2. \(\omega>\sqrt{\dfrac{g}{R}}\)
3. \(\omega=\sqrt{\dfrac{R}{g}}\)
4. \(\omega>\sqrt{gR}\)

Explanation

The equilibrium condition is

\[
\cos\theta=\dfrac{g}{\omega^2R}
\]

For a real solution,

\[
\dfrac{g}{\omega^2R}<1 \] Thus, \[ \omega>\sqrt{\dfrac{g}{R}}
\]

A particle moves in a circle of radius R. Its speed doubles while the radius becomes half. The centripetal acceleration becomes:

1. 2 times
2. 4 times
3. 8 times
4. 16 times

Explanation

\[
a_c=\dfrac{v^2}{R}
\]

New acceleration:

\[
a’_c
=
\dfrac{(2v)^2}{R/2}
=
8\dfrac{v^2}{R}
\]

Hence it becomes 8 times.

A particle moves in a vertical circle. At the side point, the string tension is \(5mg\). What is the centripetal acceleration at that point?

1. \(3g\)
2. \(4g\)
3. \(5g\)
4. \(6g\)

Explanation

At the side point, weight has no radial component.

Therefore,

\[
T=\dfrac{mv^2}{R}
\]

Given

\[
T=5mg
\]

Thus,

\[
a_c=\dfrac{v^2}{R}=5g
\]

A particle is projected horizontally with speed \(\sqrt{gR}\) from the highest point of a smooth sphere. What is the normal reaction at the instant of projection?

1. \(mg\)
2. \(\frac12mg\)
3. 0
4. \(2mg\)

Explanation

At the top,

\[
mg-N=\dfrac{mv^2}{R}
\]

Substituting

\[
v=\sqrt{gR}
\]

gives

\[
mg-N=mg
\]

Therefore,

\[
N=0
\]

The particle leaves the surface immediately.

A particle moves in a circle of radius R. The tangential acceleration is always equal in magnitude to the centripetal acceleration. If the speed at some instant is v, the angular acceleration at that instant is:

1. \(\dfrac{v^2}{R}\)
2. \(\dfrac{v}{R}\)
3. \(\dfrac{v^2}{R^2}\)
4. \(\dfrac{v^3}{R^2}\)

Explanation

Given

\[
a_t=a_c
\]

But

\[
a_t=\alpha R
\]

and

\[
a_c=\dfrac{v^2}{R}
\]

Equating,

\[
\alpha R=\dfrac{v^2}{R}
\]

\[
\alpha=\dfrac{v^2}{R^2}
\]