Uniform Circular Motion Motion in a circle at constant angular speed. ω: angular velocity (rad/s)
Rotation Angle The rotation angle is the ratio of arc length to radius of curvature. For a given angle, the greater the radius, the greater the arc length. = θ s r θ = rotation angle s = arc length r = radius
Radian Measure s = r θ = θ s r θ r i s i 2r i 2 s i
Radian Measure s = r θ = θ s r θ r i s i 2 s i 3 s i 3r i
Radian Measure s = r θ = θ s r 4 s i θ r i s i 2 s i 4r i 3 s i
How many radians in 360 0? Ans: 360 0 = 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? s =? s = C = 2πr 2 π rad = 360 0 1 rad = 360 0 /2 π = 57.3 0
How many radians in 90 0? 90 0 = 90 0 1 90 0 = 90 0 2π 360 0 Conversion Factor 90 0 = 2π 90 0 360 0 1 90 0 = 2π = π/2 4
How many radians in 22 0? 22 0 = 22 0 1 90 0 = 90 0 = 22 0 2π 2π 360 0 22 0 360 0 900 =. 38 rad
θ r = 4 m s A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of 43 0. What is the distance, d, that the box moves? d 43 = 43 x1 = 43 x 2 =.75rad π 360 First: How many radians is 43 0?
θ r = 4 m s A box is fastened to a string that is wrapped around a pulley. The pulley turns through an angle of.75 rad. What is the distance, d, that the box moves? d s = r θ d = r θ d = 4m (0.75 rad) d = 3 m rad d = 3 m 43 0 =.75rad Radian is dimensionless and is dropped!
ω = θ t s/ r = t = s/ r Angular Velocity ω: angular velocity (rad/s) t v = v r ω = θ = t r t
Rolling 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. ω = θ = t v r = = 25 m/ s.5m 50 rad / s
Angular & Tangential Velocity ω: angular velocity (rad/s) v: tangential velocity (m/s) ω = θ = t v r v = ωr 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 time BUT point 2 must travel a greater distance in the same time so it must travel at a faster tangential velocity! v = ωr
Skaters If the skaters complete one circle in 10 s, what is their average angular speed? ω = θ t = 2π 10s =.6 rad / 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 =.6 rad / s vt R = ω v = (.6 rad / s)2m T vt = 1.2 m/ s (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 =.6 rad / s vt R = ω v = (.6 rad / s)6m T vt = 3.6 m/ s 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 Force & Acceleration Center Seeking Direction: Toward the center Magnitude: 2 a c = v r a c
Angular and Tangential Acceleration a c = v r 2 at = α R α ω = t
Where is a c greatest on a merrygo-round, a or b? a c = v 2 /R but, v = Rω a b a c = (Rω) 2 /R = R 2 ω 2 /R = ω 2 R
The further out - the faster and the funner! a c = ω 2 R a b v = Rω
Translational and Rotational Kinematics v f = v o + a t x = 1/2 (v o + v f ) t ω f = ω o + α t θ = 1/2 (ω o + ω f ) t x = v o t+ 1/2 a t 2 θ = ω o t+ 1/2 α t 2 v f 2 = v o2 + 2a x ω f 2 = ω o2 + 2αθ
Translational vs Rotational m x v= x/ t a = v/ t F = ma v f = v o + a t s = v= a a c 2 I θ ω = θ / t α = ω/ t τ = Iα x = 1/2 (v o + v f )t t θ = 1/2 (ω o + ω f )t τ Connection = ω = = r ω α Fr θ r r r ω f = ω o + α t x = v o t+ 1/2 a t 2 θ = ω o t+ 1/2 α t 2 v f 2 = v o2 + 2a x ω f 2 = ω o2 + 2αθ
Angular Tangential A bike wheel with a radius of 0.25 m undergoes a constant angular acceleration of 2.50 rad/s 2. 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 = 2.5 rad / s d) How far did the bike travel? t = 4s 1 2 1 2 2 θ = ω0t+ αt = 5 rad / s4s + 2.5 rad / s (4 s) = 40rad 2 2 2 ω = ω + αt = 5 rad / s + 2.5 rad / s 4s = 15 rad / s f 0 v = ωr = 15 rad / s.25m = 3.75 m / s d = s = θ r = 40 rad.25m = 10m 2
Rotational Inertia The resistance of an object to rotate. The further away the mass is from the axis of rotation, the greater the rotational inertia.
Rotational Inertia Depends on the distribution of mass. 2 2 11 2 2 I = mr + mr
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!!!! ICM = mr 2 ICM = 1 2 mr 2
Torque: Causes Rotations τ = Fr sinφ = Fd lever arm: d = rsinφ The moment 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
Newton s 1 st Law for Rotation If the sum of the torques is zero, the system is in rotational equilibrium. τ boy = 500N 1.5m = 750Nm τ = 0 τ girl = 250N 3m =+ 750Nm
Newton s 1 st 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 forces 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 = 24.3kg R =.314m a) τ =? τ = Fr+ Fr 1 2 = ( + 90N 125 N).314m = 11Nm
Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. τ = I α Acceleration thing Force thing Inertia thing
Net Torque m = 24.3kg a) τ = 11Nm R =.314m b) α =? = τ τ α = I 1 mr 2 11Nm = 1/2 24.3kg(.314m) 2 2 = 9.2 rad / s 2
Center of Mass 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, momentum 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 parabolic path the projectile would have followed with no explosion! If no external forces act on the system, then the velocity of the CM doesn t change!!
Center of Mass The geometric center or average location of the mass. System of Particles: x CM = i M mx i total i
A meter stick has a mass of 75.0 grams and has two masses 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 moment of inertia 150 kg m 2. If the wheel starts from rest, how long will it take the wheel to make a quarter turn (90 degrees)? 1 Use : θ = ω0t + αt 2 τ = Iα α = τ 1 2 2 θ 2 θ θ = ω0t+ αt t = = 2 α τ / I 2 I t 2 π /2rad = = 2 50 N m /150kg m 3.1s
Dart Problem Determine the force M needed to give the dart a tangential speed of 5.0m/s in 0.10s starting from rest. Ignore gravity and assume the forearm has a moment of inertia of 0.065 kgm 2 (including the dart) and that the force M acts perpendicular to the forearm. r 1 r 2 τ Fr 2 = = Iα a I r 1 F = a I rr 1 2 = I v/ t rr 1 2 5 m/ s/ 0.1s F =.065kgm 2 = 464N (.025 m)(.28 m)
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 a c?
Problem: Pulsar T =.033s x 2 =.066s R = D/2 = 500km = 5x10 5 m v T =? a c =? ω 2 = θ = π t t 2π 2π 5 7 vt = ωr= R= 5x10 m= 4.76x10 m/ s t.066s 2 7 2 T (4.76 10 / ) 5 ~ 1/10 speed of light! v x m s ac = = = 4.53x10 m/ s R 5x10 m 9 2 ~ 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 T v 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 τ = Fr sinφ = Fd τ = Fd 1 1+ Fd 2 2 = ( 20 N)(.5 m) + (35 N)(1.10msin 60) OR = ( 20 N)(.5 m) + (35Ncos30)(1.10 m) F and d must be mutually perpendicular! = + 23.3 Nm CCW