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Home , Circular motion

Uniform Circular Motion in a at constant angular .

ω: angular (rad/s) The rotation angle is the ratio of arc length to of curvature. For a given angle, the greater the radius, the greater the arc length.

∆s ∆=θ r

∆θ = rotation angle ∆s = arc length r = radius ∆s Measure ∆=θ r ∆=∆srθ

2∆si ∆s ∆θ i

ri 2ri ∆s Radian Measure ∆=θ r ∆=∆srθ

3∆si 2∆si ∆s ∆θ i

ri 3ri ∆s Radian Measure ∆=θ r ∆=∆srθ

4∆si 3∆si 2∆si ∆s ∆θ i

ri 4ri How many in 3600 ? Ans: 3600 = 2 π radians

Consider a circle with radius r. θ = s/r θ = 2 πr /r θ = 2 π

r How many degrees in 1 radian? 2 π rad = 3600 0 0 s = ? s = C = 2πr 1 rad = 360 /2 π = 57.3 How many radians in 900? 900 = 900 × 1 2π 0 = 0 90 90 × Conversion Factor 3600 900 900 = 2π × 3600 1 900 = 2π × = π/2 4 How many radians in 220? 220 = 220 × 1 2π 900 = 220 × 3600 220 900 = 2π × 3600 900 = . 38 rad A box is fastened to a string that is wrapped around a s pulley. The pulley turns θ through an angle of 430. r = 4 m What is the , d, that the box moves? 43= 43x 1 d 2π = 43 x 360 = .75rad First: How many radians is 430 ? A box is fastened to a string that is wrapped around a s pulley. The pulley turns θ through an angle of .75 rad. r = 4 m What is the distance, d, that the box moves?

s = r θ d d = r θ d = 4m (0.75 rad) d = 3 m rad

0 d = 3 m 43 = .75rad Radian is dimensionless and is dropped! ω: angular velocity (rad/s)

∆θ ω = ∆t

∆sr/ ∆t = ∆t ∆∆st/ = r v = ∆θ v r ω = = ∆tr Calculate the angular velocity of a 0.500 m radius car tire when the car travels at a constant speed of 25.0 m/s. ∆θ v ω = = ∆tr

25ms / = .5m

= 50rad / s Angular & Tangential Velocity

ω: angular velocity (rad/s) v: tangential velocity (m/s)

∆θ v ω = = ∆tr

vr= ω

ω is constant & v varies with r.

The greater the r the greater v! Tangential Velocity Point 1 & 2 travel the same angle in the same BUT point 2 must travel a greater distance in the same time so it must travel at a faster tangential velocity!

vr= ω Skaters

If the skaters complete one circle in 10 s, what is their average angular speed?

∆θ ω = ∆t 2π = 10s

= .6rad / s

Don’t drop rad for angular speed. Keep it to define units. Skaters

What is the tangential velocity of a skater that is 2 m from the axis of rotation?

ω = .6rad / s

vRT = ω

vT = (.6 rad / s )2 m

vT =1.2 ms /

(The rad unit is is dropped!) Skaters

What is the tangential velocity of a skater that is 6 m from the axis of rotation?

ω = .6rad / s

vRT = ω

vT = (.6 rad / s )6 m

vT = 3.6 ms /

The further out from the axis or rotation, the faster the tangential velocity! Tangential Velocity

Is the tangential velocity constant? Velocity is a vector! Tangential Velocity

The magnitude is constant but the direction changes! Centripetal & “Center Seeking”

Direction: Toward the center Magnitude: v2 a = c r

ac Angular and Tangential Acceleration v2 ∆ω = ac aRt = α α = r ∆t Where is ac greatest on a merry- go-round, a or b? 2 ac = v /R

but, v = Rω

a b 2 ac = (Rω) /R = R2 ω2/R = ω2R The further out - the faster and the funner!

2 ac = ω R

a b v = Rω Translational and Rotational vf = vo + a ∆t ωf = ωo + α t

∆x = 1/2 (vo + vf ) t ∆θ = 1/2 (ωo + ωf ) t

∆x = v ∆t+ 1/2 a t2 ∆θ = ω ∆t+ 1/2 α t2 o o

2 2 2 2 vf = vo + 2a ∆ x ωf = ωo + 2αθ Translational vs Rotational

m Connection I ∆x ∆θ v=∆∆ xt/ ∆=∆srθ ωθ=∆∆/ t a=∆∆ vt/ vr= ω αω=∆∆/ t F= ma 2 τα= I arc = ω v = v + a ∆t ω = ω + α t f o ar= α f o ∆x = 1/2 (vo + vf )t t ∆θ = 1/2 (ωo + ωf )t ∆ 2 ∆θ ω α 2 x = vo t+ 1/2 a t τ = Fr = o t+ 1/2 t 2 2 ⊥ 2 2 vf = vo + 2a ∆ x ωf = ωo + 2αθ Angular –Tangential A bike wheel with a radius of 0.25 m undergoes a constant of 2.50 rad/s2. The initial angular speed of the wheel is 5.00 rad/s. After 4.00 s a) What angle has the wheel turned through? Givens: b) What is the final angular speed? ω0 = 5/rad s c) What is the final tangential speed of the bike? α = 2.5rad / s2 d) How far did the bike travel? ts= 4

1 1 ∆=θωtt + α2 =+=5rad / s 4 s 2.5 rad / s22 (4 s ) 40 rad 0 2 2 2 ωf = ωα0 + t =+=5rad / s 2.5 rad / s 4 s 15 rad / s

v=ω r =15 rad / s .25 m= 3.75 m / s d=∆=∆ sθ r =40 rad .25 m = 10 m Rotational The resistance of an object to rotate. The further away the is from the axis of rotation, the greater the rotational inertia. Rotational Inertia Depends on the distribution of mass. 22 I= mr11 + mr 2 2 Rotational Inertia Depends on the axis. Which has greater Rotational Inertia? (Both have Same Mass) Moments of Inertia of Various Rigid Objects Rotational Inertia Which reaches the bottom first? (Same mass and radius) Why Solid Cylinder? It has less rotational Inertia!!!!

2 ICM = mr

1 I= mr 2 CM 2 : Causes τφ=Frsin = Fd lever arm: dr= sinφ The arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force

The horizontal component of F (F cos φ) has no tendency to produce a rotation ’s 1st Law for Rotation If the sum of the is zero, the system is in rotational equilibrium.

τ boy =500N ⋅=− 1.5 m 750 Nm

∑=τ 0

τ girl =250N ⋅=+ 3 m 750 Nm Newton’s 1st Law: Conditions for Equilibrium If the sum of the net external torques is zero, the system is in rotational equilibrium.

∑=τ 0

If the sum of the net external is zero, the system is in translational equilibrium. ΣF = 0 Torque Is there a difference in torque? (Ignore the mass of the rope)

NO! In either case, the lever arm is the same! What is it? 3m Net Torque m= 24.3 kg Rm= .314 a)?∑τ =

∑τ =Fr12 + Fr =+−( 90N 125 Nm ).314 = −11Nm Newton’s 2nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia the angular acceleration. ∑=ταI Acceleration thing

Force thing Inertia thing Net Torque m= 24.3 kg a)∑τ = − 11 Nm Rm= .314 b) α = ? ∑∑ττ α= = 1 I mR 2 2 − 11Nm 2 = 2 =− 9.2rad / s 1/2⋅ 24.3kg(.314m) The geometric ‘center’ or average location of the mass. Center of Mass: Stability

If the Center of Mass is above the base of support the object will be stable. If not, it topples over. Balance and Stability

This dancer balances en pointe by having her center of mass directly over her toes, her base of support.

https://www.youtube.com/watch?v=o9rFZ_SzGPI Slide 12-88 Rotational & Translational Motion Objects rotate about their Center of Mass.

The Center of Mass Translates as if it were a point particle. Center of Mass

The Center of Mass Translates as if it were a point particle and, if no external forces act on the system, is then conserved. This means: EVEN if the bat EXPLODED into a thousand pieces, all the pieces would move so that the momentum of the CM is conserved – that is, the CM continues in the parabolic trajectory!!!! THIS IS VERY VERY IMPORTANT! System of Particles Center of Mass

• A projectile is fired into the air and suddenly explodes • With no explosion, the projectile would follow the dotted line • After the explosion, the center of mass of the fragments still follows the dotted line, the same If no external forces parabolic path the act on the system, projectile would have then the velocity of followed with no the CM doesn’t explosion! change!! Center of Mass The geometric ‘center’ or average location of the mass.

System of Particles:

∑ mxii i xCM = M total A meter stick has a mass of 75.0 grams and has two attached to it: 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark. Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances? A 50 N m torque acts on a wheel of 150 kg m2. If the wheel starts from rest, how long will it take the wheel to make a quarter (90 degrees)? 1 Use: ∆=θω t + α t 2 0 2 τ τα=→=I α I 1 22∆∆θθ ∆θω =t + α tt2 →= = 0 2/ατI

2⋅π /2rad ts= = 3.1 50N⋅⋅ m /150 kg m2 Dart Problem

Determine the force M needed to give the dart r1 a tangential speed of 5.0m/s in 0.10s starting from rest. Ignore and assume the forearm has a moment of inertia of 0.065 kgm2 (including the dart) and that the force M acts perpendicular to the forearm. r2 τα= I a a ∆∆vt/ Fr2 = I FI= = I r1 rr12 rr12 5ms / / 0.1 s F= .065 kgm2 = 464N (.025mm )(.28 ) Problem: Pulsar

Pulsar X sweeps the Earth in a radio beam once every .033 s. If the neutron star has a diameter of 1000 km, what is the tangential velocity of a point on its surface at the equator? What is ac? Problem: Pulsar

T = .033s x 2 = .066s R = D/2 = 500km = 5x105 m

vT=? ac = ? ∆θπ2 ω = = ∆∆tt 22ππ v=ω R = R = 5 x 1057 m= 4.76 x 10 ms / T ∆ts.066 ~ 1/10 speed of light! v2 (4.76 x 1072 ms / ) =T = = 92 ac 5 4.53x 10 ms / R5 xm 10 ~ 1/2 giga-g! Can on a String a c

b

If the string breaks, the can will a) continue to move in a circle b) move away opposite to the string c) move away tangent to the circle Can on a String

v T

What keeps the can moving in a circle? (ignore gravity)

The can wants to fly away tangent to its path but the string pulls it back towards the center, keeping it in circular motion. Net external torques ∑τ Find the Net Torque τφ=Frsin = Fd

∑τ =Fd11 + Fd 2 2

=−+( 20Nm )(.5 ) (35 N )(1.10 m sin 60) F and d must be OR mutually perpendicular! =−+( 20Nm )(.5 ) (35 N cos30)(1.10 m )

= +23.3Nm CCW