Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 8 Page 1 of 35
Midterm 1: Monday October 5th 2014 Motion in one, two and three dimensions Forces and Motion I (no friction) No energy and work. Page 2 of 35
6. Forces and Motion II Dynamics of uniform circular motion Centripetal acceleration: a = v2 r r r = radius of the circle v = speed Page 3 of 35 r = r r unit radial vector
Newton s 2nd law: F net = m a F net F net = mv2 r r Centripetal Force Page 4 of 35 A centripetal force accelerates a body by changing the direction of the body s velocity without changing the body s speed.
ball attached to a string in uniform circular motion on a horizontal frictionless plane F net = T (tension) the string breaks uniform linear motion, according to Newton s 1st law. no T no centripetal force a = 0 Page 5 of 35
i-clicker A ball attached to a string rotates in a vertical plane near Earth s surface such that the string is stretched taut all points on the circular path. What is the minimum value of its speed at the highest point A of the trajectory. A A) v min = 0 B) v min = rg C) v min = rg/2 D) v min = 2gr E) none of the above Page 6 of 35
Answer A ball attached to a string rotates in a vertical plane near Earth s surface. For a point P on the trajectory, let T P denote the magnitude of the tension in the string. Which of the following statements is true? A A) v min = 0 B) v min = rg C) v min = rg/2 D) v min = 2gr E) none of the above Page 7 of 35
A F net = mv2 r r Fnet = F g + T ( Fnet ) y = (mg + T ) = mv2 F g T T = mv2 mg 0 r v 2 rg v min = rg r Page 8 of 35
Example: A ball attached to a string of length l, which makes an angle θ with the vertical, rotates uniformly in a horizontal plane as shown below. Find the speed v. v l r q T F g y q T T x Page 9 of 35
F net = m a Fnet = F g + T T y q T (F net ) x = T sin θ (F net ) y = mg + T cos θ a x = v 2 /r a y = 0 F g T x T sin θ = mv2 = mv2 r lsinθ T cos θ mg = 0 Page 10 of 35 v 2 lgsinθ = tan θ v = sinθ lg cosθ
Example: a car of mass m moving with constant speed makes a turn of radius r. Suppose the static friction coefficient between the wheels and the road is µ s. Find the maximum value of the speed v such that the car will remain on the road. Page 11 of 35 f s F N F g
f s F N F g F net = m a a = mv2 r î F N mg = ma y = 0 f s = ma x = mv2 r f s µ s F N mv2 r v µ s rg µ s mg Page 12 of 35
What is energy? 7. Kinetic energy and work Basic Idea: scalar quantity with quantifies the state of motion and/or the capacity for motion of a system Conserved as the state of motion of the system changes. Page 13 of 35
Page 14 of 35 Speed of arrow in flight? Speed of train at the lowest point?
Energy Kinetic energy : associated to motion Potential energy : associated to the capacity of generating motion Example: freely falling ball from height h. initial speed: v 0 = 0 Page 15 of 35 h final speed: v 2 = v 2 0 + 2gh = 2gh
initially: potential energy: mgh h finally: kinetic energy: mv 2 /2 Energy conservation: mgh = mv2 2 Page 16 of 35
Kinetic Energy: energy associated to the motion of an object K = 1 2 mv2 Unit: Joule Page 17 of 35 1J = 1kg m 2 /s 2
Non-zero force Acceleration: change in velocity Change in kinetic energy Page 18 of 35 How does K change under an applied force?
Work: energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work while energy transferred from the object is negative work. Page 19 of 35
How do we calculate the work of a force? Example: bead on frictionless rod subject to constant force F Newton s 2nd law: F x = ma x Constant acceleration model: v 2 x = v 2 0x + 2a x x Page 20 of 35 mv 2 2 mv2 0 2 = F x x
Work done by the force F F dcos φ = F d d = ( x)î displacement vector F dcos φ > 0, for 0 φ < π/2 positive work, K f > K i = 0, for φ = π/2 zero work, K f = K i < 0, for π/2 < φ π negative work, K f < K i Page 21 of 35
Net work: When two or more forces act on an object, the net work done on the object is the sum of the works done by the individual forces. W net = W = F d = ( F ) d = Fnet d Page 22 of 35
Work-Kinetic Energy Theorem K = W K f = K i + W Page 23 of 35
i-clicker An object of mass m = 1 kg is launched with initial speed v 0 = 2 m/s along a rough horizontal surface. What is the total work done by the frictional force until it stops? A) 2 J v 0 B) 1 J C) 1 J D) 2 J E) 0 J. Page 24 of 35
Answer An object of mass m = 1 kg is launched with initial speed v 0 = 2 m/s along a rough horizontal surface. What is the total work done by the frictional force until it stops? A) 2 J v 0 B) 1 J C) 1 J D) 2 J E) 0 J. W = K = K f K i = mv2 0 2 Page 25 of 35
i-clicker A ball of mass m attached to string is launched with initial speed v 0 on a circular trajectory on horizontal plane. The friction force between the ball and the surface has constant magnitude f k. What is the speed of the ball after travelling a distance s along the circle. v s v 0 A) v = v 0 B) v = v0 2 2f ks m C) v = v 0 f ks mv 0 D) cannot be determined from the data. Page 26 of 35
Answer A ball of mass m attached to string is launched with initial speed v 0 on a circular trajectory on horizontal plane. The friction force between the ball and the surface has constant magnitude f k. What is the speed of the ball after travelling a distance s along the circle. v s v 0 A) v = v 0 B) v = v0 2 2f ks m C) v = v 0 f ks mv 0 D) cannot be determined from the data. Page 27 of 35
Infinitesimal displacement ds = vdt Note that T ds while f k makes an angle θ = 180 with ds. Infinitesimal work: Total work: dw fk = f k ds = f k ds W = Work-energy theorem: mv 2 dw = f k s 2 mv2 0 2 = f ks v = v0 2 2f ks m Page 28 of 35
Work done by the gravitational force d F g v 1 Freely falling object moving downwards: F g v 2 W = F g d = mgdcos 0 = mgd > 0 Page 29 of 35 K 1 = mv2 1 2 K 2 = mv2 2 2 K 2 K 1 = mgd > 0
d F g v 2 Freely falling object moving upwards: F g v 1 W = F g d = mgdcos 180 = mgd > 0 K 1 = mv2 1 2 K 2 = mv2 2 2 K 2 K 1 = mgd < 0 Page 30 of 35
i-clicker An object of mass m is launched with initial speed v 0 along an inclined plane making an angle θ = 45 with the horizontal. The kinetic friction coefficient between the object and the plane is µ k = 0.5. Let W fk be the total work done by the friction force until it stops. Which of the following statements is false? Page 31 of 35 v 0 q A) W fk < 0 B) W fk = mv 2 0 /2 C) W fk < mv 2 0 /2
Answer An object of mass m is launched with initial speed v 0 along an inclined plane making an angle θ = 45 with the horizontal. The kinetic friction coefficient between the object and the plane is µ k = 0.5. Let W fk be the total work done by the friction force until it stops. Which of the following statements is false? Page 32 of 35 v 0 q A) W fk < 0 B) W fk = mv 2 0 /2 C) W fk < mv 2 0 /2
y FN F gx F N q x f k q F g f k F g F g y F net = m a (F net ) x = ma x (F net ) y = ma y = 0 Page 33 of 35 F net = F g + F N + f k (F net ) x = mgsinθ f k (F nety ) y = F N mgcosθ = 0 F N = mgcosθ f k = µ k mgcosθ
f k F N f k = µ k mgcosθ d q h F g h = dsinθ Work done by kinetic friction: W fk = f k d = f k d = µ k mgdcosθ Work done by gravitational force: W Fg = F g d = mgh = mgdsinθ Work done by normal force Page 34 of 35 F N d = 0
f k F N Total work: W = W fk +W Fg = mgd ( sinθ+µ k cosθ ) d q h F g Work-Kinetic energy theorem: W = K = 0 K i = mv 2 0/2 Work done by kinetic friction: W fk W = µ kcosθ sinθ + µ k cosθ = 1 3 Page 35 of 35 W fk = mv2 0 6