Further problems on linear and angular velocity

1500 60π =50 rad/s The linear velocity of a point on the rim, v=ωr, where r is the radius of the wheel, i.e.

540 2 mm=

0.54

2 m=0.27 m.

Thus linear velocity v=ωr=(50)(0.27) =13.5 m/s

Problem 15. A car is travelling at 64.8 km/h and has wheels of diameter 600 mm.

(a) Find the angular velocity of the wheels in both rad/s and rev/min.

(b) If the speed remains constant for 1.44 km, determine the number of revolutions made by the wheel, assuming no slipping occurs.

(a) Linear velocityv=64.8 km/h =64.8km h ×1000 m km × 1 3600 h s =18 m/s. The radius of a wheel= 600

2 =300 mm =0.3 m.

From equation (5),v=ωr, from which, angular velocityω= v

r =

18 0.3 =60 rad/s

From equation (4), angular velocity,ω = 2πn, where n is in rev/s.

Hence angular speed n= ω 2π = 60 2πrev/s =60× 60 2πrev/min =573 rev/min (b) From equation (1), sincev=s/t then the time

taken to travel 1.44 km, i.e., 1440 m at a constant speed of 18 m/s is given by:

time t= s

v =

1440 m 18 m/s =80 s

Since a wheel is rotating at 573 rev/min, then in 80/60 minutes it makes

573 rev/min× 80

60min=764 revolutions Now try the following exercise.

Exercise 67 Further problems on linear and angular velocity

1. A pulley driving a belt has a diameter of 300 mm and is turning at 2700/π revolu- tions per minute. Find the angular velocity of the pulley and the linear velocity of the belt assuming that no slip occurs.

[ω=90 rad/s,v=13.5 m/s] 2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm. Determine the linear velocity of a point on the rim of one of the wheels of the bicycle, and the angular velocity of the wheels.

[v=10 m/s,ω=40 rad/s] 3. A train is travelling at 108 km/h and has

(a) Determine the angular velocity of the wheels in both rad/s and rev/min. (b) If the speed remains constant for 2.70 km,

determine the number of revolutions made by a wheel, assuming no slipping occurs.

(a) 75 rad/s, 716.2 rev/min (b) 1074 revs

14.7

Centripetal force

When an object moves in a circular path at constant speed, its direction of motion is continually changing and hence its velocity (which depends on both mag- nitude and direction) is also continually changing. Since acceleration is the (change in velocity)/(time taken), the object has an acceleration. Let the object be moving with a constant angular velocity ofωand a tangential velocity of magnitudevand let the change of velocity for a small change of angle of θ(=ωt) be V in Fig. 14.12. Then v2−v1=V . The vector diagram is shown in Fig. 14.12(b) and since the mag- nitudes ofv1andv2are the same, i.e.v, the vector diagram is an isosceles triangle.

Figure 14.12

Bisecting the angle betweenv2andv1gives: sinθ 2 = V/2 v2 = V 2v i.e. V =2vsinθ 2 (1) Sinceθ=ωt then t= θ ω (2)

Dividing equation (1) by equation (2) gives: V t = 2vsin (θ/2) (θ/ω) = vωsin (θ/2) (θ/2) For small angles sin (θ/2)

(θ/2) ≈1, hence V t = change of velocity change of time =acceleration a=vω However, ω= v r (from Section 14.6) thus vω=v· v r = v2 r i.e. the acceleration a is v

2

r and is towards the cen- tre of the circle of motion (along V ). It is called the centripetal acceleration. If the mass of the rotating object is m, then by Newton’s second law, the cen- tripetal force ismv

2

r and its direction is towards the

centre of the circle of motion.

Problem 16. A vehicle of mass 750 kg travels around a bend of radius 150 m, at 50.4 km/h. Determine the centripetal force acting on the vehicle.

The centripetal force is given by mv 2

r and its direction is towards the centre of the circle.

Mass m=750 kg,v=50.4 km/h = 50.4×1000

60×60 m/s =14 m/s

and radius r=150 m,

thus centripetal force=750(14) 2

B

Problem 17. An object is suspended by a thread 250 mm long and both object and thread move in a horizontal circle with a constant angu- lar velocity of 2.0 rad/s. If the tension in the thread is 12.5 N, determine the mass of the object.

Centripetal force (i.e. tension in thread),

F= mv

2

r =12.5 N

Angular velocityω=2.0 rad/s and radius r =250 mm=0.25 m.

Since linear velocityv=ωr,v=(2.0)(0.25) =0.5 m/s. Since F=mv 2 r , then mass m= Fr v2, i.e. mass of object, m=(12.5)(0.25)

0.52 =12.5 kg Problem 18. An aircraft is turning at constant altitude, the turn following the arc of a circle of radius 1.5 km. If the maximum allowable accel- eration of the aircraft is 2.5 g, determine the maximum speed of the turn in km/h. Take g as 9.8 m/s2.

The acceleration of an object turning in a circle is

v2

r . Thus, to determine the maximum speed of turn,

v2 r =2.5 g, from which, velocity,v=(2.5gr)=(2.5)(9.8)(1500) =√36750=191.7 m/s and 191.7 m/s=191.7×60×60 1000 km/h=690 km/h Now try the following exercise.

Exercise 68 Further problems on cen-