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.). \[ F_c = m a_c = \frac{m v^{2}}{r}= m\omega^{2}r\qquad(\text{units N}) \]

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 \(m a_c\). 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 \(\vec a_c\). Because the direction of \(\vec a_c\) rotates with the object, it is a true vector quantity, not a scalar “force”.

Derivation of the Acceleration Formula

1. Two successive velocity vectors \(\vec v_1\) and \(\vec v_2\) are separated by a small angle \(\Delta\theta\). The magnitude of their difference is \[ \Delta v = |\Delta\vec v| = 2v\sin\frac{\Delta\theta}{2}\;\approx\;v\Delta\theta\quad(\Delta\theta\ll1). \]
2. The average acceleration over \(\Delta t\) is \[ a_{\text{avg}} = \frac{\Delta v}{\Delta t}\approx\frac{v\Delta\theta}{\Delta t}=v\omega . \]
3. 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

Common Misconceptions

1. “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.).
2. “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.
3. “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.
4. “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).
5. “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.

1. Calculate the centripetal acceleration: \[ a_c = \frac{v^{2}}{r}= \frac{(4.0)^2}{0.80}=20\ \text{m·s}^{-2}. \]
2. Apply \(F_c = m a_c\): \[ F_c = (0.50)(20)=10\ \text{N}. \]
3. 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.

1. Resolve the normal reaction \(N\) into components: \[ N\cos\beta = mg,\qquad N\sin\beta = \frac{m v^{2}}{r}. \]
2. Divide the second equation by the first to eliminate \(N\): \[ \tan\beta = \frac{v^{2}}{r g}\;\Longrightarrow\; v = \sqrt{r g \tan\beta}. \]
3. 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.

1. At the top, weight and tension both point toward the centre: \[ m\frac{v^{2}}{r}= mg + T_{\text{min}}. \]
2. 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).

1. Rotate the mass at a steady rate and measure the period \(T\) of one revolution with the stopwatch.
2. Calculate angular speed \(\omega = 2\pi/T\).
3. Predict the required centripetal force: \(F_{\text{pred}} = m\omega^{2}r\).
4. Measure the tension in the string directly with the spring balance (\(F_{\text{exp}}\)).
5. 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: \(F_c = m a_c\).
• 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.

Suggested Diagram (Insert in the margin)