understand that centripetal acceleration causes circular motion with a constant angular speed
Centripetal Acceleration – A‑Level Physics (Cambridge 9702)
Learning Objective
Explain why a centripetal (radial) acceleration is required for an object to move in a circle with a constant angular speed, relate it to Newton’s 2nd law, and use the relevant equations to solve quantitative problems, including experimental verification.
Key Definitions & Relationships
- Radian (rad) – the angle subtended at the centre of a circle by an arc whose length equals the radius.
\(2\pi\;\text{rad}=360^{\circ}\). - Angular displacement \(\theta\) – measured in radians; linear arc length \(s = r\theta\).
- Angular speed \(\omega\) – rate of change of angular displacement.
\[
\omega = \frac{2\pi}{T}=2\pi f =\frac{v}{r}\qquad(\text{units rad·s}^{-1})
\]
where \(T\) is period and \(f\) is frequency.
- Linear (tangential) speed \(v\) – speed along the circular path.
\[
v = \omega r\qquad(\text{units m·s}^{-1})
\]
- Centripetal (radial) acceleration \(a_c\) – always directed toward the centre of the circle.
\[
a_c = \frac{v^{2}}{r}= \omega^{2}r = v\omega\qquad(\text{units m·s}^{-2})
\]
- Centripetal force \(F_c\) – the net radial force that produces the required acceleration (tension, friction, component of weight, normal reaction, etc.).
\[
Fc = m ac = \frac{m v^{2}}{r}= m\omega^{2}r\qquad(\text{units N})
\]
Link to Newton’s 2nd Law & Work Done
- Newton’s 2nd law in vector form, \(\displaystyle \mathbf{F}{\text{net}} = m\mathbf{a}\), tells us that the *net* radial force acting on a body moving in a circle must equal \(m ac\). The term “centripetal force” therefore does not denote a new kind of force; it is simply the vector sum of all real forces that have a component toward the centre.
- The centripetal force is always perpendicular to the instantaneous displacement (\(\mathbf{v}\) is tangent, \(\mathbf{F}_c\) is radial). Because work \(W = \mathbf{F}\!\cdot\!\mathbf{s}\) involves the component of force along the displacement, the centripetal force does no work on the object; it changes direction of motion but not its kinetic energy.
Vector Nature of Centripetal Acceleration
The velocity vector \(\vec v\) is tangent to the circular path. In a short interval \(\Delta t\) the direction changes by \(\Delta\theta =\omega\Delta t\) while the magnitude stays constant. The change in velocity \(\Delta\vec v\) points toward the centre, giving an inward acceleration \(\vec ac\). Because the direction of \(\vec ac\) rotates with the object, it is a true vector quantity, not a scalar “force”.
Derivation of the Acceleration Formula
- Two successive velocity vectors \(\vec v1\) and \(\vec v2\) are separated by a small angle \(\Delta\theta\). The magnitude of their difference is
\[
\Delta v = |\Delta\vec v| = 2v\sin\frac{\Delta\theta}{2}\;\approx\;v\Delta\theta\quad(\Delta\theta\ll1).
\]
- The average acceleration over \(\Delta t\) is
\[
a_{\text{avg}} = \frac{\Delta v}{\Delta t}\approx\frac{v\Delta\theta}{\Delta t}=v\omega .
\]
- Taking the limit \(\Delta t\to0\) gives the instantaneous centripetal acceleration
\[
\boxed{a_c = v\omega = \frac{v^{2}}{r}= \omega^{2}r }.
\]
Important Formulas
| Quantity | Symbol | Formula | Units |
|---|---|---|---|
| Angular displacement | \(\theta\) | \(s = r\theta\) | rad |
| Angular speed | \(\omega\) | \(\displaystyle\omega = \frac{2\pi}{T}=2\pi f = \frac{v}{r}\) | rad·s\(^{-1}\) |
| Linear speed | v | \(v = \omega r\) | m·s\(^{-1}\) |
| Centripetal acceleration | ac | \(ac = \frac{v^{2}}{r}= \omega^{2}r = v\omega\) | m·s\(^{-2}\) |
| Centripetal force | Fc | \(Fc = m a_c = \frac{m v^{2}}{r}= m\omega^{2}r\) | N |
Common Misconceptions
- “Centripetal force is a new kind of force.” It is the *net* radial force – the vector sum of existing forces (tension, friction, component of weight, normal reaction, etc.).
- “If speed is constant, there is no acceleration.” Acceleration also includes a change in direction. In uniform circular motion the speed is constant but the velocity direction changes continuously, giving a non‑zero acceleration.
- “Centrifugal force is a real outward force.” In an inertial frame there is no outward force; the feeling of being “pushed out” is a fictitious force that appears only in a rotating (non‑inertial) reference frame.
- “The centripetal force must be supplied by a single contact.” The required radial force can be the resultant of several forces acting together (e.g., tension plus a component of weight on a banked curve).
- “The centripetal force does work.” Because it is always perpendicular to the instantaneous displacement, the work done by the centripetal force is zero.
Worked Example 1 – Horizontal Whirling
Problem: A 0.50 kg ball is attached to a light string and whirled in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m·s\(^{-1}\). Find the tension in the string.
- Calculate the centripetal acceleration:
\[
a_c = \frac{v^{2}}{r}= \frac{(4.0)^2}{0.80}=20\ \text{m·s}^{-2}.
\]
- Apply \(Fc = m ac\):
\[
F_c = (0.50)(20)=10\ \text{N}.
\]
- In a horizontal plane the only radial force is the tension, so
\[
T = 10\ \text{N}.
\]
Worked Example 2 – Banked Curve (No Friction)
Problem: A car of mass 1200 kg travels round a banked circular road of radius 50 m, bank angle \(\beta = 15^{\circ}\). Find the speed for which no friction is required.
- Resolve the normal reaction \(N\) into components:
\[
N\cos\beta = mg,\qquad N\sin\beta = \frac{m v^{2}}{r}.
\]
- Divide the second equation by the first to eliminate \(N\):
\[
\tan\beta = \frac{v^{2}}{r g}\;\Longrightarrow\; v = \sqrt{r g \tan\beta}.
\]
- Insert numbers:
\[
v = \sqrt{(50)(9.8)\tan15^{\circ}} \approx 12.0\ \text{m·s}^{-1}.
\]
Worked Example 3 – Vertical Circle (Tension at Top)
Problem: A 0.20 kg ball is attached to a light string of length 0.60 m and swung in a vertical circle. Find the minimum speed at the top of the circle so that the string remains taut.
- At the top, weight and tension both point toward the centre:
\[
m\frac{v^{2}}{r}= mg + T_{\text{min}}.
\]
- For the string just to stay taut, \(T_{\text{min}} = 0\). Hence
\[
\frac{v^{2}}{r}=g\;\Longrightarrow\; v = \sqrt{gr}= \sqrt{9.8\times0.60}\approx 2.4\ \text{m·s}^{-1}.
\]
Experimental Verification of \(a_c = \omega^{2}r\)
Apparatus: low‑friction turntable, small mass \(m\) on a string of known length \(r\), stopwatch, spring balance (or force sensor).
- Rotate the mass at a steady rate and measure the period \(T\) of one revolution with the stopwatch.
- Calculate angular speed \(\omega = 2\pi/T\).
- Predict the required centripetal force: \(F_{\text{pred}} = m\omega^{2}r\).
- Measure the tension in the string directly with the spring balance (\(F_{\text{exp}}\)).
- Compare \(F{\text{pred}}\) and \(F{\text{exp}}\); agreement within experimental uncertainty confirms the relationship \(a_c = \omega^{2}r\).
Summary Checklist
- Define a radian and convert to degrees.
- Remember \(\displaystyle \omega = \frac{2\pi}{T}=2\pi f = \frac{v}{r}\) (rad s\(^{-1}\)).
- Use the vector diagram to see why \(\vec a_c\) points inward.
- Apply any of the three equivalent forms of centripetal acceleration:
\[
a_c = \frac{v^{2}}{r}= \omega^{2}r = v\omega .
\]
- Link the required centripetal force to Newton’s 2nd law: \(Fc = m ac\).
- Recall that the centripetal force does no work on the object.
- Identify the real forces that combine to give the net radial force in each situation (tension, friction, component of weight, normal reaction, etc.).
- Check for common misconceptions and explicitly correct them in exam answers.