Physics Department Tutorial: Motion in a Circle (solutions)

Transcription

1 JJ 014 H Physics (9646) o Solution Mark 1 (a) The radian is the angle subtended by an arc length equal to the radius of the circle. Angular elocity ω of a body is the rate of change of its angular displacement. (For a rigid rotating body, its angular elocity is the same for all points of the body.) P Sentence in brackets are optional. Physics Department Tutorial: Motion in a Circle (solutions) (c) 1. The Earth orbits around the Sun in 1 year = 365 days. Angular elocity of Earth around the Sun, π π ω = = T (365x4x3600) =.0 x 10-7 rad s -1. The Earth rotates about its axis in 1 day = 4 hours. π π Angular elocity of Earth about its axis, ω = = T (4x3600) = 7.3 x 10-5 rad s -1 3 (a) Angle QOP = π /3 rad (since ΔOPQ is an equilateral triangle). π Therefore, angular displacement of OQ relatie to OP = π 3 5π = rad Ans: ( D ) = = ( π ) ω = 16 rad s -1 t Linear speed of the rim, = rω = (16) =.0 m s -1 4 Tension in the cord, T = Mg Using F x = ma x = mrω, Mg sin = mrω Mg sin = m (Lsin)ω Mg = m L ω ω = (Mg / ml) Mg mg T π π Lm ( / M) = = = π ω Mg g ml Physics Dept Page 1 of 5

2 JJ 014 H Physics (9646) 5 = rω and ω number of reolution per unit time. Since ω is constant r. Graph A is correct. Ans: ( A ) 6 (a) The direction of its elocity is changing continuously. As its elocity is changing, the body must hae an acceleration. Consider the body moing through a small angular displacement from point P to point Q along a circular path as shown in Fig. 7.1 below. P Q Change in elocity Final elocity - initial elocity Fig. 7.1 Fig. 7. The change in elocity Δ, always points towards the centre of the Δ circular path as shown in Fig. 7.. Since acceleration, a =, the Δ t direction of the acceleration should also point in the same direction as Δ which is towards the centre of the circle. -ans Alternatie explanation: Since the direction of elocity is always changing, without a change in the speed, we conclude that there cannot be any component of acceleration tangential to the circle. There can only be a radial acceleration, i.e. along the radial line. As the object needs to moe inwards in a circular path, this implies that the acceleration must also be radially inwards, hence towards the centre of the circle. 7 (a) Since the angular displacement (0.010 rad) is small, the magnitude of the change in elocity Δ = x = rad Δ Direction of the acceleration of P is towards the centre of the circle. - (c) The arc length, s traelled by the object = r (where is the angular displacement) Time taken for P to trael through is gien by Δ t = s r r = = Physics Dept Page of 5

3 JJ 014 H Physics (9646) (d) Magnitude of the acceleration of P is gien by Δ a = = = Δt ( r / ) r 8 As the ball traels in the horizontal circle, the tension in the string proides the centripetal force pointing towards the centre of circle. Since the force on ball by string (which is the tension) is towards the centre of circle (i.e. towards left in the diagram), by ewton s Third Law of Motion, the pull of ball on string must be in the opposite direction (i.e. towards right in the diagram). Hence, the pull of pole on string at the other end of the string must be in the opposite direction to that of pull of ball on string (i.e. towards left in the diagram). Ans: ( D ) 9 To keep cylinder at rest, T = Mg. This tension proides the centripetal force, T = Mg = m r = Mgr / m 10 Since the car is in equilibrium in the ertical direction, F y = m a y = 0 R cos = mg (1) Since there is a centripetal acceleration towards the centre of the circular motion, F C = m a C = m R /r R sin = m /r () Equations ()/(1): tan = /gr = tan -1 ( /gr) = tan -1 {13.4 /(9.81 x 50.0)} = 0.1 mg 11 (a) ω = π / T = π / 3. = rads -1 F C = R A - mg mrω = R A - mg R A = mrω + mg = 63(6.6)(1.964) + 63(9.81) = 1.08 = 0 R A F C = R B + mg mrω = R B + mg R B = mrω - mg = 63(6.6)(1.964) - 63(9.81) = = 985 R B Physics Dept Page 3 of 5

4 JJ 014 H Physics (9646) c) In order to maintain the speed constant at this position, the weight has to be counteracted by an upward force. So that there is no resultant force along the ertical tangent to the circular path. This upward force is the frictional force, f, exerted by the seat on the rider. The normal contact force,, exerted by the seat on the rider will proide for the centripetal force. f f = = mg = 63 x 9.81 =618 = mrω = 63(6.6)(1.964) = = 1600 Force exerted by the seat on the rider = f + = 618 = 1715 = f Force exerted by the seat on rider (ote: eight of rider is not exerted by the seat) 1 At the minimum speed, the bicycle just starts to lose contact with the ground = 0. m Centripetal force, F C =mg + = mg + 0 = R m min = mgr = gr = 5.15 ms -1 m R Since the equation aboe is independent of mass, m, the minimum speed will not change een if Diaolo used a heaier bicycle. 13 (a) A ector quantity has both magnitude and direction, while a scalar quantity only has magnitude. Acceleration is the rate of change of elocity of an object. Physics Dept Page 4 of 5

5 JJ 014 H Physics (9646) (c) (i) Change in elocity from A to B = Final elocity Initial elocity = Final elocity + ( Initial elocity) Change in elocity, Δ Final Final Velocity elocity - Initial elocity Hence, Δ = = 5.5 m s -1 (ii) Acceleration of the car between A and B d Δ 5.5 = = = = 5.8 m s - dt t 4.4 (c) (iii) Resultant Force Centre of Circle (i) The resultant force acts perpendicularly to the direction of motion of the car. This causes a change in the direction of motion only (i.e. elocity changes, implying there is an acceleration), without any change in the speed of the car. Hence, the kinetic energy of the car remains unchanged. -ans Physics Dept Page 5 of 5