PY105 (C1)

1. Assignment 5 on circular motion and impulse will be due next Monday. Uniform Circular 2. Lecture notes can be downloaded from the Motion following URL: http://physics.bu.edu/~okctsui/PY105.html

Note that it is case-sensitive. You can also access this website through WebCT.

1 2

DEFINITION OF UNIFORM CIRCULAR MOTION Four common ways to express the speed Uniform circular motion is the motion of an object traveling a of a uniform circular motion constant speed on a circular path. 1. Specify the (uniform) speed, v. 2. Specify the period, τ , to complete one revolution. 3. Specify the angular speed, ω. r 4. Specify the number of revolutions per unit .

r Notice that assumed in this definition is the circular path (with a v given radius, r) that defines the trajectory of the motion. Also notice that the velocity, v, at any time is along the tangent of the circular path at where the object is. 3 4

Four common ways to express the speed Example 1: A Tire-Balancing Machine of a uniform circular motion The wheel of a car has a radius of 0.29m, and is rotating (1) speed, v, (2) period, , (3) angular speed, τ ω at 830 revolutions per minute on a tire-balancing and (4) number of revolutions per unit time machine. Determine the speed at which the outer edge We can relate v and τ by the following reasoning: of the wheel is moving. Distance traveled in one revolution = 2πr 1 =1.2×10−3 min revolution Hence, vτ = 2πr 830revolutions min

r τ =1.2×10−3 min = 0.072 s ⇒τ= 2πr/v or v = 2πr/τ v 2π r 2π (0.29 m) Secondly, ω = 2π/τ = v/r or v = rω v = = = 25m s τ 0.072 s Lastly, number of revolutions per unit time = 1/τ 5 6

1 Centripetal Acceleration v(t ) ∆v(t )= v(t +∆t)+(−v(t0)) 0 The fact that the direction of motion of an object in a circular 0 0 (180ο −θ)/2 = 90ο −θ/2 motion changes continuously with time suggests that the 90ο −θ/2 velocity vector of the object varies continuously with time. θ θ v(t +∆t) Hence the motion must have an acceleration, a. We call −v(t0) 0 θ accelerations giving rise to circular motions centripetal Drawing from accelerations. last page:

o o ∆v(t0) v(t0) r What is the centripetal r at t0 θ 90ο −θ/2 v at t acceleration, a at a v at t0 c θ p 0 c θ p c θ θ given time, t0? θ/2 r at t0+∆t v(t0+∆t) v at t+∆t v at t0+∆t n.b. |v(t +∆t)|=|v(t )| Note that v, being tangential to the circular path, is 0 0 perpendicular to the radial direction. 7 8 See appendix I for a detailed explanation.

Centripetal Acceleration v (a) (b) ∆ B A r(t0) θ v(t ) r θ v(t0+∆t) ∆v(t ) 0 −v(t ) 0 c θ 0 θ 90ο −θ/2 r(t0+∆t) c θ θ θ/2 Circle drawn by using v as the radius and the of –v(t0) as the center. v(t0+∆t) From (a), v∆t = rθ⇒∆t = rθ/v In the limit ∆t → 0, θ→0. n.b. |v(t0+∆t)|=|v(t0)|=v From (b), when θ→0, ∆v = vθ Acceleration, a = lim∆t →0∆v/∆t. This drawing shows vθ v2 a = lim ∆v/∆t = = that ∆v and hence a points towards the center. ∆t →0 r 9 rθ/v 10

Example 2: The Effect of Radius on Centripetal Acceleration Centripetal Force The racing track contains turns with radii of 33 m and 24 m. the centripetal acceleration In circular motions, the object is subject to the at each turn for a speed of 34 m/s. Express centripetal acceleration, a . So, a net force must act 2 c answers as multiples of g = 9.8 m/s . on an object to produce a circular motion. We call Solution: the component of the net force that produces the

2 circular motion the centripetal force, FC. Accordingly, Use ac = v r FC = mac. Notice that the magnitude of Fc is ()34m s 2 determined entirely by ma = mv2/r. r = 33 m: a = = 35m s2 = 3.6g c c r 33 m F = mar = mar + mar 2 ∑ c other ()34m s 2 r = 24 m: ac = = 48m s = 4.9g 24 m This term is non-zero Fc if the net force is Note that the centripetal acceleration required to 11 12 a turn is bigger the sharper the turn is. bigger than Fc

2 Centripetal Force (Conceptual) Exercise 3 Object on a turntable Object For an object placed on top of ω In Example 3, the centripetal a rotating turntable, what is forces, F , required by the f the source of the centripetal vr c S r racing car in making the turns fS force on the object if it rotates are provided by the static with the turntable? fS frictional forces between the Ans. Static friction road and the tires of the wheels. Turntable If the radial position, r, of the object fS’ What will happen if the static on the turntable is doubled, how friction is not large enough to fS’ much does the centripetal force provide the centripetal force change? f ’ needed for making the turn? S 2 2 2 2 Ans. Fc = /r = m(2πr/τ) /r = 4π mr/τ 13 14 Hence, if r is doubled, Fc will be doubled.

Example 4: Centripetal Force Caused by a Tension (Conceptual) Exercise 3 Object on a turntable (cont’d) The model airplane has a mass of 0.90 kg and moves at a constant speed of 19 If static friction is not enough to provide the m/s on a circle that is parallel to the centripetal force required by the object to ground. The path of the airplane and the rotate with the turntable, which way will the guideline lie in the same horizontal plane object go? because the weight of the plane is balanced by Object the lift generated by its wings. Find the tension Ans. It will follow the straight ω in the 17 m guideline.

path along which the velocity vr Solution: FBD at an instant: vector, v , is pointed. r 2 r v L (Lift from Fc = T = m a the wings) r T (19m s)2 Turntable T = ()0.90 kg =19 N 15 17 m mg 16

Example 5: Centripetal Force by a Banked Curve Example 5: Centripetal Force by a Banked Curve (cont’d) On a banked curve that makes an angle θ with the horizontal, the centripetal force is mostly provided by the horizontal component of the normal force. The vertical component of the normal force balances the car’s weight. Ignore friction, what is the relation between θ and v?

From the above, we deduce that the car remains v 2 in the circular path if = tan θ ⇒ v = rg tan θ . rg Solution: v 2 What will happen if the driver increases the speed 2 1/2 FN sinθ = ma = m v of the car from v = (rgtanθ) ? r tanθ = rg 17 18 FN cosθ − mg = 0

3 Example 5: Centripetal Force by a Banked Curve (cont’d) Vertical Circular Motions

fsinθ f Motions confined to a circular fcosθ path but under gravitational acceleration, g. At different points of the circular path, mg contributes to the centripetal But if friction is not negligible and the condition, tanθ = v2/rg is not satisfied (i.e., v2 ≠ rgtanθ), static friction comes into force, Fc, differently, so the play to make the net force equal to the required centripetal normal force, FN, which force = mv2/r. provides the difference 2 v 2 x: F sinθ − f cosθ = m Note that if v > rgtanθ, between Fc and mg varies N r f < 0 meaning that f with position. y: FN cosθ + f sinθ − mg = 0 points towards the mg v 2 center to help provide ⇒ f = (tanθ − ) 19 20 sinθ tanθ + cosθ rg the centripetal force.

Consider the vertical circular motion If the speed is uniform, as shown. FN3

of mg to ac: If FN = 0, the cyclist loses contact with the track. Amongst all positions it 2 v1 Maximum happens most probably at 3 . 1 F − mg = m 2 N1 r negative v contribution FN3 + mg = m v2 r 2 F = m 2 F = 0 ⇒ v = (rg)1/2 N 2 r No N3 contribut- 2 If v is less than (rg)1/2, F = 0 and part of mg would v ion N 4 F = m 4 cause the object to fall at 3 : N 4 r mg = mv2/r + ma v2 Maximum y,fall 3 F + mg = m 3 positive N 3 ΣF may,total r contribution 21 22

Another example of vertical circular motions

Consider an object tied to a string v3 and being maneuvered into a T vertical circular motion as shown at mg right. The criterion for the object to ac remain in the circular trajectory is that the tension in the string r remains non-zero throughout the

motion. As before, if speed v3 at the highest point is 2 such that mv3 /r < mg, tension T will be zero and the difference, mg – mv2/r, will cause the object to fall vertically and leave the circular path.

23

4