141. PROP. A point moves with constant speed v velos along the perimeter of a regular polygon ABC... of n sides inscribed in a circle of radius r ft.; prove that at each angle the point receives an additional velocity towards the centre of the circle ; vo and that the rate of additional velocity per second is Let O be the centre of the circle. Let AB=s ft.; let 7 seconds be the interval occupied in going from A to B; then, s = 2'7. r Draw ON bisecting AB at right angles. Produce AB to K so that BK represents the velocity v velos along AB; let BH represent the velocity v velos along BC; then BH=BK. But the angle ABL = BKH; therefore the angle ABL = HBL. Therefore BL produced passes through the centre O. Now the velocity represented by BH is the resultant of the velocities BK, BL. That is, the velocity BL when added to the velocity BK changes it into the velocity BH. In other words, when a point having v velos along AB changes its velocity to v velos along BC, it receives an additional velocity w velos represented by BL, whose direction is towards the centre of the circle. The rate of additional velocity per second is T S 316 31 316 = HK 2KW 2NB and, V ВК BK OB v2 S vo x X 71? 1 Q. E. D. 142. PROP. A point moves with the constant speed v velos along the circumference of a circle of radius r; prove that this point has constant acceleration celos towards the centre of the circle. We consider a circle to be the curve which a regular polygon approaches as its limit, when the number of its sides is infinitely increased. When the polygon in Art. 141 becomes a circle the additional velocity at each angular point is infinitely diminished, the number of angular points being infinitely increased. Hence, instead of a series of sudden increases of velocity we have, for the circle, a continuous growth of velocity, va always in the direction of the centre, at the rate va second; that is, at the rate N.B. Neither in the polygon nor in the circle does this continuous increase of velocity alter the speed. It alters the velocity as regards direction only. velos per 1 r 143. PROP. A particle of mass m lbs. is describing a circle of radius r ft. with v velos; prove that it is acted on by mv? a force poundals towards the centre. 22 The particle has a constant acceleration celos towards the centre. Acceleration never exists. unless force causes it; hence mv there is a force poundals towards the centre. 144. A force which always tends towards a fixed point is called a centripetal force. Example i. A particle of 1 lb., fastened to a fixed point O by a light string 2 ft. long, describes the circle about O as centre once every second; find the tension of the string. [NOTE. In a question like this, gravity is not supposed to be acting. The particle might be moving on a perfectly smooth horizontal plane. The student should notice that unless some word such as, weight, horizontal, vertical, heavy, appears in a question, gravity is supposed not to act.] The particle describes the circumference of a circle of 2 ft. radius in 1 sec. ; therefore it moves at the rate of 47. ft. per sec. 1672 Therefore by Art. 142 its acceleration is celos. Therefore it is acted on by 872 poundals. The tension of the string is therefore equal to the weight of about 8x22 X 22 pounds, nearly ; 32 X 49 that is, to the weight of 2.47 lbs. nearly. NOTE. When a mass is describing a circle with uniform speed, its velocity is being changed' (not in magnitude, but) in direction. By reason of its inertia. this change cannot be made in the velocity of the mass by anything except by external force. We have shewn that in order to effect this change a force of a certain magnitude must continuously act on the particle towards the centre of the circle. The inertia of the mass continuously opposes this force; and used to be said to be centrifugal. This word centrifugal is very misleading ; for the student should carefully notice that if the centripetal force ceases, the motion of the mass is along a tangent: therefore the tendency of the mass should be called tangential, not centrifugal. 2 EXAMPLES. XXXV. 1. A point is describing a circle of radius 5 ft. with uniform speed 10 velos, find its acceleration. 2. A mass of 4 lbs. is describing a circle of radius 5 ft. with uniform speed of 10 velos; what force is acting upon it? 3. A mass m lbs. fastened by a string to a fixed point is describing a circle uniformly with v velos; the tension of the string is p poundals, find the length of the string. 4. A mass m lbs. is fastened by a string of length r feet to a fixed point O, and describes a circle about on times per second; shew that the tension of the string is 2mnarar poundals. 5. A railway carriage is describing a curve on horizontal rails ; shew that if it goes fast enough it must fall over. 6. A railway carriage of 10 tons is describing an arc of a circle of 4 mile radius with a speed of 30 miles an hour; find the horizontal force which must be acting on the carriage. 7. A mass of 1 cwt. is in a railway truck which is moving, on horizontal rails on a curve whose radius is 400 yards, at the rate of 60 miles an hour; what is the horizontal stress between 1 cwt. and the truck? 8. A mass of 56 lbs. is placed in a swing the ropes of which are 10 ft. long; find the difference in the tension of the ropes when the swing is at its lowest point, (i) when it is at rest, (ii) when it is moving with 20 velos (neglecting the mass of the swing itself). 9. A horse weighing 5 cwt. is going along a road at the rate of 10 miles an hour and passes over a depression in the road whose section is that of a circle of 40 feet radius; what is the difference between his weight and the average pressure of his feet on the road when at the lowest point of the depression? 10. Two masses of 3 lbs. and 4 lbs. on a smooth horizontal plane are fastened together by a light inelastic string 7 ft. long and are each describing a circle so that one point in the string is at rest on the plane; what must be the point in the string? 11. Two masses A and B of 1 lb. each are joined by a string; the mass A describes a circle of radius 2 ft. with uniform speed on the surface of a smooth horizontal table, while the other mass B is suspended under the action of gravity by the string, which passes through a small hole in the table at the centre of the circle; find the speed of A. 12. Two masses A and B are joined by a string; the mass A of 10 lbs. describes on a smooth horizontal table a circle of radius 5 ft. with uniform speed 5 velos; the other mass B is suspended under the action of gravity by the string which passes through a small hole in the table at the centre of A's circle ; find the mass of B that it may rest in equilibrium. 13. A particle of 3 lbs. is fastened to a fixed point by a string 2 N2 feet long and is describing a horizontal circle under the action of gravity, so that the string describes a right circular cone, the inclination of the string to the vertical being 45°; prove that the velocity of the particle is about 8 velos. 14. Shew that the tension of the string in Question 13 is 3 12 lbs. weight. 15. A particle of m lbs. is fastened to a fixed point by a string I feet long, and is describing a horizontal circle under the action of gravity, so that the string describes a right circular cone of vertical angle za; prove that the velocity of the particle is Nlig sina a sec a) velos. NOTE. The arrangement described in Question 15 is called a Conical Pendulum. 16. One end of a string 21 feet long is fastened to a point A on a fixed smooth vertical rod, the other to a small ring P of mass m lbs. which slides on the rod; another mass Q m' lbs. is fastened to the middle point of the string and revolves with velocity v velos in a horizontal circle so that the angle AQP is a right angle, prove that lg (m' + 2m) 22= m 145. PROP. When a point P is moving with uniform velocity, the line OP joining P to any fixed point o, traces out equal areas in equal intervals. Let the point P be moving in the straight line ABD with uniform velocity; P passes over equal distances in equal intervals ; let AB and BD be two distances passed over by Pin any two equal intervals; then, AB=BD. Therefore the area OAB=OBD. |