CAIE A-Level Physics A2 12 Motion in a Circle Questions
Practise analysing circular motion with radians, angular and linear speed, centripetal acceleration and force, direction, units and assumptions.
- Syllabus
- 2028–2030
- Course
- Physics 9702
- Level
- A2
Practise analysing circular motion with radians, angular and linear speed, centripetal acceleration and force, direction, units and assumptions.
Define the radian.
angle (subtended at the centre of a circle) when arc (length) = radius
A circular metal disc spins horizontally about a vertical axis, as shown in Fig. 1.1.
Fig. 1.1 (not to scale)
A piece of modelling clay is attached to the disc.
For the instant when the piece of modelling clay is in the position shown, draw on Fig. 1.1:
an arrow, labelled A , showing the direction of the acceleration of the modelling clay.
arrow, labelled A, pointing in NW direction
The metal disc in Fig. 1.1 has a radius of 9.3 cm .
The centre of gravity of the modelling clay is 1.2 cm from the rim of the disc and moves with a speed of 0.68 ms−1.
Calculate the angular speed ω of the disc.
ω=rads−1
v=rωω=0.68/(0.093−0.012)=8.4rad s−1
Calculate the acceleration a of the centre of gravity of the modelling clay.
a= ms−2
a=v2/r or a=rω2a=0.682/(0.093−0.012) or (0.093−0.012)×8.42=5.7 m s−2
A second piece of modelling clay is attached to the disc in the position shown in Fig. 1.2.
Fig. 1.2
The second piece of modelling clay has a larger mass than the first piece.
By placing one tick (✓) in each row, complete Table 1.1 to show how the quantities indicated compare for the two pieces of modelling clay.
Table 1.1
angular speed: same for both pieces
linear speed: less for second piece than first piece
acceleration: less for second piece than first piece
With reference to velocity and acceleration, describe uniform circular motion.
constant speed or constant magnitude of velocity
acceleration (always) perpendicular to velocity
Two cars are moving around a horizontal circular track. One car follows path X and the other follows path Y , as shown in Fig. 1.1.
Fig. 1.1 (not to scale)
The radius of path X is 318 m . Path Y is parallel to, and 27 m outside, path X . Both cars have mass 790 kg . The maximum lateral (sideways) friction force F that the cars can experience without sliding is the same for both cars.
The maximum speed at which the car on path X can move around the track without sliding is 94 m s−1.
Calculate F.
F= N
F=mv2/r
or
v=rω and F=mrω2F=790×942/318=22000 N
Both cars move around the track. Each car has the maximum speed at which it can move without sliding.
Complete Table 1.1, by placing one tick in each row, to indicate how the quantities indicated for the car on path Y compare with the car on path X .
Table 1.1
centripetal acceleration: same
Marking guidance:
maximum speed: greater
time taken for one lap of the track: greater
A steel sphere of mass 0.29 kg is suspended in equilibrium from a vertical spring. The centre of the sphere is 8.5 cm from the top of the spring, as shown in Fig. 2.1.
Fig. 2.1
The sphere is now set in motion so that it is moving in a horizontal circle at constant speed, as shown in Fig. 2.2.
Fig. 2.2
The distance from the centre of the sphere to the top of the spring is now 10.8 cm .
Explain, with reference to the forces acting on the sphere, why the length of the spring in Fig. 2.2 is greater than in Fig. 2.1.
horizontal force on sphere causes centripetal acceleration
weight of sphere is (now) equal to vertical component of tension
or horizontal and vertical components (of force) (now) combine to give greater tension (in spring)
greater tension in spring so greater extension of spring
Use the information in (b) to determine the centripetal acceleration of the sphere.
centripetal acceleration =(Tsinθ)/m=(3.2×sin27∘)/0.29=5.0 m s−2
Calculate the period of the circular motion of the sphere.
period =
a=rω2 and ω=2π/T
or
a=v2/r and v=2πr/TT=2π×(0.049/5.0)=0.62 s
Define the radian.
angle (subtended at centre of circle) when arc length = radius
The minute hand of a clock revolves at constant angular speed around the face of the clock, completing one revolution every hour. A small piece of modelling clay is attached to the hand with its centre of gravity at a distance L from the fixed end of the hand, as shown in Fig. 1.1.
direction of revolution of minute hand
Calculate the angular speed ω of the minute hand.
ω=2π/T=2π/(1.0×60×60)=1.7×10−3rad s−1
During a time interval of 1400 s , the centre of gravity of the piece of modelling clay in Fig. 1.1 moves through a total distance of 0.44 m .
Calculate the angle through which the minute hand moves in this time interval.
angle =1.7×10−3×1400=2.4rad
Determine distance L.
L= arc length / angle =0.44/2.4
or
L=0.44×(3600/1400)/2πL=0.18 m
Calculate the magnitude of the centripetal acceleration of the piece of modelling clay.
centripetal acceleration = ms−2
a=rω2=0.18×(1.745×10−3)2=5.5×10−7 m s−2
Use your answer in (c)(iii) to explain why the variation with time of the magnitude of the force exerted by the minute hand on the piece of modelling clay is negligible as the minute hand undergoes one full revolution.
centripetal acceleration is negligible compared with acceleration of free fall or
numerical comparison establishing answer to (c)(iii) << 9.81
resultant force is negligible compared with weight (of modelling clay) (so variation is negligible)
or
force exerted by minute hand (approximately) equal (and opposite) to weight of modelling clay