Centripetal Acceleration Questions

Centripetal acceleration is the acceleration an object feels when it moves in a circle at a steady speed. Even though the speed doesn’t change, the direction of the object’s velocity is always changing, and that change in direction means the object is accelerating. This acceleration always points toward the center of the circle, which is why it’s called centripetal, meaning "center-seeking". It’s important to know that centripetal acceleration isn’t caused by a new kind of force. Instead, it happens because of a force—called the centripetal force—that pulls or pushes the object toward the center of the circle. This force can come from things like tension in a string, gravity, or friction, depending on the situation.

1.0Acceleration in Uniform Circular Motion

Uniform circular motion: If a particle is moving in a circle with constant speed then motion is called Uniform circular motion (UCM).  

$\overrightarrow{\Delta v}={\vec{v}}_B-{\vec{v}}_A$

Magnitude of change in velocity

$|\overrightarrow{\Delta v}|=|{\vec{v}}_B-{\vec{v}}_A|=\sqrt{v_B^2+v_A^2+2v_A{v}_B\cos (\pi -\theta )}$

$$\begin{gathered} v_A=v_B=v,Sincespeedisthesame \\ |\overrightarrow{\Delta v}|=2v\sin \frac{\theta }{2} \\ DistancetravelledbyparticlebetweenAandB=r\theta \end{gathered}$$

Hence time taken, $\Delta t=\frac{r\theta }{v}$

Average acceleration,

$|\vec{a}|=\left|\frac{\Delta \vec{v}}{\Delta t}\right|=\frac{2v\sin \frac{\theta }{2}}{\frac{r\theta }{v}}=\frac{v^2}{r}\cdot \frac{2\sin \frac{\theta }{2}}{\theta }$

If $\Delta t\to 0$, then $\theta$ is small, $\sin \left(\frac{\theta }{2}\right)\approx \frac{\theta }{2}$

${\lim }_{\Delta t\to 0}\left|\frac{\Delta \vec{v}}{\Delta t}\right|=\left|\frac{d\vec{v}}{dt}\right|=\frac{v^2}{r}$

instantaneous acceleration is, $\frac{v^2}{r}$

$a=a_c=\frac{v^2}{r}$

Though derived for constant speed, the centripetal acceleration formula still applies when speed varies.

Note:

1. This acceleration is due to the change in direction of velocity and is called centripetal or normal acceleration (aₙ).
2. It always acts perpendicular to velocity, directed towards the centre.
3. In uniform circular motion (UCM), its magnitude is constant, but direction continuously changes.

2.0Non – Uniform Circular Motion

If speed varies, the motion is called non-uniform circular motion. It involves two accelerations: tangential (due to changing speed) and centripetal (due to changing direction).

$a_{\text{net}}=\sqrt{(a_t{)}^2+(a_n{)}^2}$

$\tan \theta =\frac{a_t}{a_n}$

$S=\theta R$

$$\begin{gathered} Differentiation|V|=\omega R \\ Differentiation\frac{d|V|}{dt}=R\frac{d\omega }{dt} \end{gathered}$$

$a_t=\alpha R$

1. Tangential acceleration : The acceleration component directed along the tangent to a circular path is called tangential acceleration. It is responsible for changing the particle’s speed and is defined as

$a_t=\frac{dv}{dt}=\frac{d|\vec{v}|}{dt}=\text{Rate of change of speed}$

$a_t=\alpha R$

Important Point:

• In vector form ${\vec{a}}_t=\vec{\alpha }\times \vec{r}$
• When the tangential acceleration acts in the same direction as the velocity, the particle’s speed increases.
• Conversely, if the tangential acceleration is directed opposite to the velocity, the particle’s speed decreases.

2. Centripetal acceleration : Centripetal acceleration is responsible for the change in the direction of velocity. In circular motion, this acceleration is always present, pointing toward the center of the circle. Although its magnitude may remain constant, centripetal acceleration continuously changes direction as the particle moves along the path. It is also known as radial acceleration or normal acceleration.
3. Net acceleration : The total acceleration of a particle moving along a curved path is the vector sum of its centripetal (or normal) acceleration and its tangential acceleration.

$\vec{a}=\frac{d\vec{v}}{dt}={\vec{a}}_r+{\vec{a}}_t$

$a=\sqrt{a_t^2+a_r^2}$

$\tan \theta =\frac{a_t}{a_r}$

Important Points:

(i) The derivative of speed gives tangential acceleration.

(ii) Derivative of velocity $(\vec{v})$gives total acceleration.

(iii) $\left|\frac{d\vec{v}}{dt}\right|$ is the magnitude of total acceleration, while $\frac{d|\vec{v}|}{dt}$ gives tangential acceleration. They are not the same.

3.0Acceleration of Particle in Curvilinear Motion

Examine a particle moving along a curved path. Let the net acceleration $a_{\text{net}}$ make an angle θ with the velocity ($\vec{v}$) at point P.

Then, $(a_{\text{net}})=\frac{d\vec{v}}{dt}$

• The component of $a_{\text{net}}$ along the direction of velocity is the tangential acceleration (aₜ), which changes the speed: $a_t=\frac{d|\vec{v}|}{dt}$
• The component perpendicular to velocity is the normal (centripetal) acceleration (aₙ), which changes the direction of motion, aₙ always points toward the concave side of the path.

4.0Radius of curvature

Any curved path can be considered as a collection of infinitely small circular arcs. The radius of curvature at a given point on the curve is defined as the radius of the circular arc that best approximates the curve at that specific point.

$R=\frac{v^2}{a_n}$

5.0FAQ on Centripetal Acceleration Questions

Q-1. Find centripetal acceleration of given points (A and B) as shown in figure.

Solution:

$(a_c{)}_A={\omega }^{2}R=5\times 5\times 1=25m/sec2$

$(a_c{)}_B={\omega }^{2}R=5\times 5\times 2=50m/sec2$

Q-2.A particle is moving in a circle of radius 10 cm with uniform speed completing the circle in 4s, find the magnitude of its acceleration. A particle moves with uniform speed along a circular path of radius 10 cm, completing one full revolution in 4s.Determine the magnitude of its acceleration.

Solution:

The distance covered in completing the circle is $2\pi r=2\pi \times 10cm$

The linear speed is $v=\frac{2\pi r}{t}=\frac{2\pi \times 10cm}{4s}=5\pi cm/s$

Magnitude of acceleration is $a=\frac{v^2}{r}=\frac{(5\pi cm/s{)}^2}{10cm}=2.5{\pi }^{2}cm/s^2$

Q-3.A particle is moving along a circular path of radius 2.0 cm, with its speed varying according to the equation v = 4t,  where v is in cm/s and t is in seconds.

(a) Determine the tangential acceleration of the particle at  t = 1s.
(b) Calculate the total acceleration of the particle at t = 1s.

Solution:

(a) Tangential Acceleration

$a_t=\frac{dv}{dt}\text{or}{a}_t=\frac{d}{dt}(4t)=4{\text{cm/s}}^2$

$a_c=\frac{v^2}{R}=\frac{(4t{)}^2}{2}=8{\text{cm/s}}^2$

(b)Total Acceleration

$a=\sqrt{a_t^2+a_r^2}=\sqrt{(4{)}^2+(8{)}^2}=4\sqrt{5}{\text{cm/s}}^2$

Q-4.A particle moves along a circular path with a constant tangential acceleration of $0.6m/s^2$ . It begins to slip when the magnitude of its total acceleration reaches $1m/s^2$. Determine the angle (in radians) through which the particle turns before slipping occurs.

Solution:

$a_{\text{Net}}=\sqrt{a_t^2+a_r^2}$

${\omega }^2={\omega }_0^2+2\alpha \theta (\because {\omega }_0=0)$

${\omega }^2=2\alpha \theta$

$a_c={\omega }^{2}R=2(\alpha \theta R)=2a_t\theta$

$1=\sqrt{0.36+(1.20{)}^2}\Rightarrow 1-0.36=(1.20{)}^2$

$\Rightarrow \frac{0.8}{1.2}=\theta \Rightarrow \theta =\frac{2}{3}\text{Radian}$

Q-5.A particle starts from rest and moves along a circular path with a constant angular acceleration of $4rad/se{c}^2$. Determine the time at which the magnitudes of its centripetal and tangential accelerations become equal.

Solution:

$a_t=\alpha R$

$\Rightarrow V=0+\alpha Rt\Rightarrow a_c=\frac{v^2}{R}=\frac{{\alpha }^2{R}^2{t}^2}{R}$

$a_t=a_c$

$\alpha R=\frac{{\alpha }^2{R}^2{t}^2}{R}\Rightarrow t^2=\frac{1}{\alpha }=\frac{1}{4}\Rightarrow t=\frac{1}{2}\text{sec}$

Q-6.A bee travels along a circular path (projectile motion) with constant speed of 25m/s and having a radius 160 m. Find an bee at the highest point ?

Solution:

Radius of curvature = 160m

$R=\frac{V^2}{a_n}$

$160=\frac{25^2}{a_n}\Rightarrow a_n=\frac{625}{160}m/s^2$

Note: The radius of curvature is the property of the curve and not the motion of the particle.

Q-7.A 2 kg body is placed on a smooth horizontal surface and attached to a 3 m long string. It is rotated in a horizontal circle at 60 revolutions per minute (rpm). What is its centripetal acceleration?

Solution: $m=2kg,r=3m$

$\omega =60\text{rev/minute}=\frac{60\times 2\pi }{60}rad/sec=2\pi rad/sec$

Because the angle described during 1 revolution is $2\pi$ radian.

$v=r\omega =2\pi \times 3m/s=6\pi m/s$

$a_c={\omega }^{2}r=(2\pi {)}^2\times 3=12{\pi }^2=118.4m/s^2$

Q-8.A particle is revolving in an annular path of radius 500 m at a speed 30 m/s. It is increasing its speed at the rate of $2m/s^2$. What is its acceleration? 

Solution:

$r=500\mathrm{m},u=30m/s,a_t=2{m/s}^2$

$a_c=\frac{v^2}{r}=\frac{30^2}{500}=\frac{9}{5}{m/s}^2$

Total Acceleration:

$a=\sqrt{a_t^2+a_c^2}=\sqrt{2^2+{\left(\frac{9}{5}\right)}^2}=\sqrt{\frac{181}{25}}{m/s}^2$

Q-9.A particle is moving in a loop of radius 50 cm in such a way that, at every moment, the magnitudes of its normal and tangential accelerations are equal. If it speed at t=0 is 4m/s, time taken to complete the first revolution will be $\frac{1}{\alpha }[1-e^{-2\pi }]\mathrm{s}$ where = $\alpha$

Solution:

$|{\vec{a}}_c|=|{\vec{a}}_t|$

$\frac{v^2}{r}=\frac{dv}{dt}$

${\int }_4^v\frac{dv}{v^2}={\int }_0^t\frac{dt}{r}$

$\Rightarrow {\left[-\frac{1}{v}\right]}_4^v=\frac{t}{r}$

$\Rightarrow -\frac{1}{v}+\frac{1}{4}=2t$

$\Rightarrow v=\frac{4}{1-8t}=\frac{ds}{dt}$

$4{\int }_0^t\frac{dt}{1-8t}={\int }_0^{s}ds$

$(r=50\text{cm}=0.5\text{m})$; $(s=2\pi r=\pi )$

$4\times \frac{{\left[\ln (1-8t)\right]}_0^t}{-8}=\pi$

$\ln (1-8t)=-2\pi$

$1-8t=e^{-2\pi }$

$t=\frac{1-e^{-2\pi }}{8}\text{s}$

$\therefore \alpha =8$

Q-10.A man is cycling smoothly along a circular track of radius 9 m with a monkey on his shoulder. He completes 120 revolutions in 3 minutes. What is the magnitude of the monkey’s centripetal acceleration?

Solution:

$\theta =120\times 2\pi ,t=3\text{min}=180\text{s}$

$\omega =\frac{\theta }{t}=\frac{240\pi }{180}=\frac{4\pi }{3}\text{rad/s}$

$a_c={\omega }^{2}R={\left(\frac{4\pi }{3}\right)}^2\times 9=16{\pi }^2{\text{m/s}}^2$

Q-11. A stone tied to a 180cm long string at its end is making 28 revolutions in a horizontal circle every minute. The magnitude of acceleration of stone is $1936x\frac{m}{s^2}$.The value of x is

Solution:

$a={\omega }^{2}r$

$a={\left(\frac{28\times 2\pi }{60}\right)}^2\times 1.8$

$a=\frac{1936\times 1.8}{225}=\frac{1936}{125}{\text{m/s}}^2$

$x=125$