Rotational mechanics for Cambridge International AS & A Level Physics

Reading cue. W is drawn from the dot at her centre of mass and F from the surface meeting the air; v is a velocity, not a third force, and W is drawn longer than F for a reason.

1. aDeduce Deduce from the relative lengths of the arrows W and F in Fig. 7.1 whether the skydiver has yet reached terminal velocity.

   recall2 marks

   Check answer · 2 marks

   1. F is drawn shorter than W, so there is a resultant force downwards
   2. she is therefore still accelerating and has not reached terminal velocity

2. bCalculate At the instant drawn, the drag force F is 3.4 × 10² N. Calculate the weight of the skydiver and her acceleration at that instant.

   routine3 marks

   Check answer · 3 marks

   1. W = 85 × 9.81 = 8.3 × 10² N
   2. resultant force = 834 − 340 = 4.9 × 10² N downwards
   3. a = 494 / 85 = 5.8 m s⁻² downwards

3. cDetermine The skydiver falls 250 m from rest and is then moving at 50 m s⁻¹. Determine the average drag force acting on her over that fall.

   demanding4 marks

   Check answer · 4 marks

   1. loss of gravitational potential energy = 85 × 9.81 × 250 = 2.08 × 10⁵ J
   2. gain in kinetic energy = ½ × 85 × 50² = 1.06 × 10⁵ J
   3. work done against drag = 2.08 × 10⁵ − 1.06 × 10⁵ = 1.02 × 10⁵ J
   4. average drag force = 1.02 × 10⁵ / 250 = 4.1 × 10² N

4. dExplain The skydiver is falling at constant velocity when she turns into a head-down dive, presenting a much smaller area to the airflow. Explain how the two arrows of Fig. 7.1 change, and describe her subsequent motion.

   top of the paper4 marks

   Check answer · 4 marks

   1. the drag is reduced, so F becomes shorter while W is unchanged
   2. there is now a resultant downward force, so she accelerates again
   3. as her speed rises the drag increases until F is once more equal to W
   4. she then falls at constant velocity at a higher terminal speed than before

Reading cue. The dashed line labelled r runs from the dot at the Earth's centre to the satellite, so it is the orbit radius, not the height above the surface, and it is not to scale.

1. aState State what provides the centripetal force on the satellite in Fig. 7.1, and state its direction on the figure.

   recall2 marks

   Check answer · 2 marks

   1. the gravitational attraction of the Earth on the satellite
   2. directed along the dashed line towards the dot at the centre of the Earth

2. bCalculate The orbit radius r is 4.2 × 10⁷ m and the radius of the Earth is 6.4 × 10⁶ m. Calculate the height of the satellite above the Earth's surface.

   routine2 marks

   Check answer · 2 marks

   1. height = r − radius of the Earth = 4.2 × 10⁷ − 6.4 × 10⁶
   2. = 3.6 × 10⁷ m

3. cDetermine The satellite has a mass of 1.5 × 10³ kg, and for the Earth GM = 3.99 × 10¹⁴ m³ s⁻². Determine the gravitational field strength of the Earth at this orbit radius and the force the Earth exerts on the satellite.

   demanding3 marks

   Check answer · 3 marks

   1. g = GM/r² = 3.99 × 10¹⁴ / (4.2 × 10⁷)²
   2. g = 0.23 N kg⁻¹
   3. F = mg = 1.5 × 10³ × 0.226 = 3.4 × 10² N

4. dDeduce A second satellite is placed in a circular orbit of radius 2.1 × 10⁷ m about the same centre, with GM = 3.99 × 10¹⁴ m³ s⁻² as before. Deduce its period, and deduce whether it can remain above one point on the equator.

   top of the paper4 marks

   Check answer · 4 marks

   1. GMm/r² = 4π²mr/T², so T = 2π√(r³/GM)
   2. T = 2π√((2.1 × 10⁷)³ / 3.99 × 10¹⁴) = 3.0 × 10⁴ s
   3. = 8.4 hours, so it circles the Earth almost three times each day
   4. its period is not 24 hours, so it cannot stay above a single point on the equator