This page contains the AQA AS P17 Motion In Circle Questions and kerboodle answers for revision and understanding.This page also contains the link to the notes and video for the revision of this topic.
Banner 1
C17.1 Uniform circular motion AQA A2 Physics P17 Motion In Circle Kerboodle Answers: Page No. 275
1.Ans-a clock in: a 1 second 1.75 x 10-3 rad b 1 minute 0.105 rad c 1 hour. 6.28 rad 2.Ans- a its time period, 20 ms b the angle it turns through in radians in i 1 ms, 0.31 rad ii Is. 310 rad 3.Ans- a. the speed of rotation of a point on the equator, 465 m s-1 b the angle the Earth turns through in 1 s in i degrees, 0.0042° ii radians. The radius of the Earth 6400 km. 7.3 x 10-5 rad 4.Ans- a its speed, 7.0 km s-1 b its angular displacement in 1.0 sin i degrees, 0.050° ii radians.- 7 x 10-4 rad
17.2 Centripetal acceleration AQA A2 Physics P17 Motion In Circle Kerboodle Answers : Page No. 277
1.Ans- a the speed of a capsule, 0.23 m s-1 b i) the acceleration of a capsule, the centripetal force on a person of mass 6B kg in a capsule. 7.9 x 10-4 m s-2 2.Ans- a the speed and acceleration of the object, 5 .1 x 10-2 N b the centripetal force on the 0.53 m s-1, 0.66 m s-2 3.Ans-i the speed 9.9 x 10-2 N ii the centripetal acceleration of the Earth on around the Sun. i the speed, 3.0 x 104 m s-1 ii the time for one of satellite, Radius of the Earth 5400 km Acceleration of free fall ms-2 6.0 x 10-3 m s-2 4.ANS-a the Speed Of the h before it Was released, 8.4 m s-1 b its centripetal acceleration, 88 m s-2 c the centripetal force the hammer just before it was released. 175 N17.3 On the road AQA A2 Physics P17 Motion In Circle Kerboodle Answers : Page No. 279
1.Ans- a the centripetal acceleration of the vehicle on the bridge, 6.7m s-2 b the support force On the vehicle when it was at the top. 3.8 kN 2.Ans- a the centripetal acceleration, 4.1 m s-2 b the centripetal force on the vehicle when moving at this speed. 3.0 kN 3.Ans-The banking angle needs to be greater as the speed increases. The banking of the track means that the normal reaction of the track pushing up on the sprinter helps with his/her centripetal force as it has a component towards the centre of the circle. A flat track just pushes upwards and the sprinter needs to rely more on friction to get the centripetal force. 4.Ans- a Use the equation yr tan 9 to cälculäte the speed of a vehicle On the bend if there is to be no sideways friction on its tures. 40 m s-1 b When turning the wheels, the car still wants to continue straight ahead. Because of the inertia, the car is not just stopped but wants to go on forward (the effect is known as the centrifugal effect.) When you turn the wheels, so they don’t follow the motion perfectly anymore, they should slide over the asphalt as the car continues forward. You have then caused a velocity-component perpendicular to the wheels, in which direction they can’t turn – they can only slide. But the car doesn’t start sliding and burning your Goodyear rubber tires (it doesn’t continue forward without change). Static friction will prevent that. That static friction is pulling in the wheel to oppose that slide (preventing any perpendicular velocity-component) to prevent the car from slipping and sliding. Be reminded that friction is always something that tries to prevent a motion. It will always act the exact opposite way as the motion/velocity it is trying to stop – in this case, exactly opposite to the perpendicular velocity-component of the wheel. This introduced static friction will always be perpendicularly to the wheels direction. Any parallel component of this force would have been in the direction of the wheels rotation, and so wouldn’t stop or change the motion (but just keeps the wheels rotating). And acceleration always happens in the direction of the (net) force, F⃗ =ma⃗ .17.4 At the fairground AQA A2 Physics P17 Motion In Circle Kerboodle Answers : Page No. 281
1.Ans-Practice questions: Page No. 282-285
1.Ans- (a) Each spring holds its brake pad retainer on the shaft at low speed. If the rotation speed is increased, the brake pad retainer moves away from the shaft and compresses the spring, which acts against the outward movement of the retainer. If the rotation speed is fast enough, the spring is unable to prevent the brake pad coming into contact with the collar. Friction between the brake pad and the collar prevents the shaft rotating any faster. (b) For no braking, the centripetal force < 250 N. ∴ mω02 r = 250 N at the maximum angular speed ω, 0.30 ω0 2 × 0.060 = 250 (c) If the springs became weaker, the tension in the springs at which the brake pads touched the collar would be less……so braking would occur at a lower rotation frequency. The lifeboat would descent at a lower speed, or more friction occurs 2 (a) Speed is the magnitude of velocity (or speed is a scalar but acceleration (or velocity) is a vector).- In circular motion at constant speed the direction of motion changes continuously.
- Therefore the velocity is changing.
- Acceleration is the rate of change of velocity.
- because, when the rate of rotation is constant, force ∝ radius r (F = mω2r and ω is constant).
- Friction (or electrostatic attraction) is sufficient to hold the dust particles that are closer to the centre but not those further away.
- Stress is
- F depends on the mass of the engine, the speed of the engine, and the radius of the track.
- A is the area of contact between the wheel and the rail.
- A discussion of how changing a physical quantity would affect the stress, for example increasing the mass of the engine would increase the stress, or an increase in the depth of the flange would decrease the stress.
- Tension, arrow along thread towards centre of circle.
- Air resistance (or drag), arrow along a tangent to the circle in the opposite direction to the rotation arrow.
- The movement of the pointer is to the left (or away from the centre).
- The right hand spring must stretch to provide this force.