(O Level Phy) A question on speed-time graph and force in a circular motion

Here's a question modified from May/June 2007 Paper 3 Extended.

A model car moves around a horizontal circular track. (a) A force acts on the car to keep it moving in a circle. In which direction does the force act? [1] The speed of the car increases. State and explain what happens to the magnitude of this force. [2] (b) The car travels too quickly and leaves the track at a point on the track. In which direction will the model car travel after it has left the track? Explain your answer. [2] In terms of the forces acting on the car, suggest why it left the track at the point. [2] (c) The car, starting from rest, increases its speed uniformly for 3.0 s. After that it travels at a constant speed of 25 m/s. It completes one lap of track in 10 s. Sketch a speed-time graph for the motion described. [2] Use your graph to calculate the circumference of the track. [2] Calculate the increase in speed per second during the first 3.0 s. [2]

You need to grasp a few concepts before you can answer the question.

• Kinematics: Speed-time graphs give us numerous information including the acceleration (calculated from the gradient) and distance travelled (calculated from the area under the graph). Don't use distance = speed x time which is for constant speed. Always include units in your answers. If the speed is in m/s while the time is in s, acceleration would be in m/s² while distance would be in m.
  
• Dynamics: An object moving in a circular motion at a constant speed would have a changing velocity due to its changing direction. Since the object in circular motion changes velocity constantly, it is experiencing an acceleration. This acceleration points towards the centre of the circular motion. We know, from Newton's 2^nd Law (∑F = ma), that the net force and acceleration are in the same direction. Therefore, there's a net force acting in the centre of the circular motion. Note: If this net force is lost suddenly, the object's inertia would cause it to move in a straight line along the tangent of the circular path.
  

The gradient of a speed-time graph for a motion in a straight line is the acceleration...

The area under the speed-time graph is the distance covered...

An object moving in a circular motion experiences a force and an acceleration pointing towards the centre of the circular path...

Now, try the above question before you check with the answer below. Be careful of the units as the speed is in cm/s and so the distance travelled would be in cm and the rate of change in speed would be in cm/s².

(a)(i) The force acts towards the centre of the circular track. [1]
(ii) The force increases in magnitude. [1] This is because a larger force is needed to keep the model car in the circular track. [1]

(b)(i) The model car travels in a straight line along the tangent of the circular track. [1] This is because the inertia of the moving car keeps it traveling in a constant direction. [1]
(ii) The friction between the car's wheels and the track provides the resultant (net) force to keep the model car from leaving the circular track. [1] When the car moves too fast, its wheels lose grip with the track. [1]

(c)(i) [See diagram below]



(ii) Circumference
= area A + area B
= ½ x 25 cm/s x 3.0 s + 25 cm/s x (10.0 – 3.0) s [1]
= 212.5 cm [1]
(ii) Increase in speed per second in the first 3.0 s
= gradient of speed-time graph in the first 3.0 s
= (25 cm/s – 0) / 3.0 s [1]
= 8.33 cm/s² [1]

Note that the question in (c)(iii) writes “increase in speed per second” instead of “acceleration”. This is because the car is moving in a circular motion, not in a straight line.