1 Motion of a single particle - Linear momentum, work and energy principle

Transcription

1 1 Motion of a single particle - Linear momentum, work and energy principle 1.1 In-class problem A block of mass m slides down a frictionless incline (see Fig.). The block is released at height h above the bottom of the loop. (a ) What is the force of the inclined track on the block at the bottom (point A)? Find an expression in terms of h and R. (b ) What is the force of the track on the block at point B? (c ) At what speed does the block leave the track? (d ) How far away from point A does the block land on level ground? Solution (a) I. Choose the reference system Consider the reference system as shown in the figure of the assignment. The free-body diagram at point A: Applying linear momentum principle in the y direction: P y = F y mv /R = N mg. (1.1) Remark: make sure you re-read example I.3 in the lecture notes. Since all forces are either potential (G = mg) or do no work (N), the system is conservative, thus, we can use conservation of energy principle. Total energy at the top of the track: E top = T top + V top = 0 + mgh. (1.) Total energy at point A: E A = T A + V A = 1 mv + 0. (1.3) 1-1

2 1 Motion of a single particle - Linear momentum, work and energy principle 1- Now we can calculate v from: E top = E A mgh = 1 mv v = gh. (1.4) Substituting (1.4) into (1.1) gives: ( N = mg 1 + h ). R (1.5) (b)i. Choose the reference system Consider the reference system as shown in the figure of the assignment. The free-body diagram at point B: Applying linear momentum principle in the normal direction at point B. P n = F n mv = N mg cos 45 R The force of the track exerted on the block at point B is the following: N = mv R + mg. (1.6) We know that the total energy according to (1.) is E tot = mgh. Since the system is still conservative, get v by the conservation of energy. The total energy at point B is: E B = T B + V B = 1 mv + mgh B, (1.7) where h B can be written as: ( h B = R 1 1 ). (1.8)

3 1 Motion of a single particle - Linear momentum, work and energy principle 1-3 Finally the conservation equation E tot = E B reads mgh = 1 ( mv + mgr 1 1 ). (1.9) Solving equation (1.9) for v gives: [ ( gh gr 1 1 )] = v, (1.10) and substituting (1.10) into (1.6) yields [ ( )] h 3 N = mg R +. (1.11) (c) From (1.10): [ ( v = gh gr 1 1 )]. (1.1) (d) Projectile motion I. Choose the reference system Considering point A as the reference system origin. The free-body diagram. Applying linear momentum principle in the x and y direction: P x = F x mẍ = 0 and P y = F y mÿ = mg Integrating twice with respect to time these equations and substituting the initial conditions, namely x(0) = x 0, y(0) = y 0, v x (0) = v x0, v y (0) = v y0, we get the equation of motion for the projectile. x = x 0 + v x0 t, (1.13)

4 1 Motion of a single particle - Linear momentum, work and energy principle 1-4 y = y 0 + v y0 t 1 gt, (1.14) where g = 9.81 m/s. Substituting the already known quantities (1.13) and (1.14) became x = R + v B t, (1.15) y = h B + v B t 1 gt. (1.16) Now solve (1.16) for t when y = 0 (the block reaches the ground) gt h B v B t = 0, t = vb ± vb + 8gh B. (1.17) g We discard the negative root since it gives a negative time and substitute into (1.15): ) x = R + v B ( vb + v B + 8gh B g Using the previous expressions for v B and h B yields x = ( 1)R + h +. (1.18) h 3 R + R. (1.19)

5 1 Motion of a single particle - Linear momentum, work and energy principle In-class problem A bead, under the influence of gravity, slides along a frictionless wire whose height is given by the function y(x). Assume that at position (x, y) = (0, 0), the wire is vertical and the bead passes this point with a given speed v 0 downward. What should the shape of the wire be (y as a function of x) so that the vertical speed remains v 0 at all times? Solution I. Choose the reference system Consider the reference system as shown in the figure of the assignment. Free-body diagram: Since all forces are either potential (G = mg) or do no work (N), the system is conservative, thus, we can use conservation of energy principle. The bead s speed at any time is given by 1 mv + mgy = 1 mv 0 v = (Note that y is negative here). The vertical component of the velocity is v 0 gy. (1.0) ẏ = dy dt = v sin θ, where the angle θ is defined as shown below at every point of the curve. (1.1)

6 1 Motion of a single particle - Linear momentum, work and energy principle 1-6 Therefore sin θ = y / 1 + y with y = dy/dx being the slope of the wire. Now the requirement is ẏ = v 0, which is equivalent to v sin θ = v 0. (1.) Then we get the following formula by substituting (1.0) into (1.) ( ) v0 y gy = v 0. (1.3) 1 + y Squaring both sides and solving for y = dy/dx yields dy dx = v 0 gy. Separating variables and integrating gives gy dy = v0 dx ( gy)3/ 3g (1.4) = v 0 x, (1.5) where the constant of integration has been set to zero, because (x; y) = (0; 0) is a point on the curve. Therefore the shape of the wire is: y = (3gv 0x) /3 g. (1.6)

7 1 Motion of a single particle - Linear momentum, work and energy principle In-class problem The ball of mass m is given a speed of v A = 3gr at position A. When it reaches B, the cord hits the small peg P, after which the ball describes a smaller circular path. Determine the position x of P so that the ball will just be able to reach point C. Solution I. Choose the reference system Normal-tangential (n-t) coordinates are attached to, and move with, the ball. Free-body diagram: Applying linear momentum principle in the normal direction at point C: P n = F n m v C r x = mg v C = g(r x) (1.7) (Note: when the ball is just about to complete the small circular path, the cord becomes slack at position C, i.e., T = 0, a n = v C.) r x Since all forces are either potential (mg) or do no work (T ), the system is conservative, thus, we can use conservation of energy principle. With reference given in Fig.(b), the gravitational potential energy of the ball at positions A and C are V A = mgh A = 0, V C = mgh C = mg(r x). (1.8) (Note: do not confuse v with V.)

8 1 Motion of a single particle - Linear momentum, work and energy principle 1-8 Conservation of Energy: T A + V A = T C + V C 1 m(3gr) = 1 mv C + mg(r x) (1.9) v C = g(x r) (1.30) Solving (1.7) and (1.30) yields: x = r and v 1 3 C = gr. 3

9 1 Motion of a single particle - Linear momentum, work and energy principle Homework The m = 65 kg skier starts from rest at A. Assuming that the friction, air resistance and the size of the skier can be neglected: (a ) Determine his speed at point B as a function of h AB. (b ) Determine the distance l and the time t when he lands in C. (c ) Calculate the distance l and the time t when he lands in C, if α = 30, h AB = 15 m and h BD = 4.5 m. Solution I. Choose the reference system Consider the reference system as shown in the figure of the assignment. Free-body diagram at point B: (a) Since all forces are either potential (G = mg) or do no work (N), the system is conservative, thus, we can use conservation of energy principle (v A = 0 and y B = y 0 ). 1 mv A + mg(y A y 0 ) = 1 mv B + mg(y B y 0 ) v B = g(y A y B ) = gh AB (1.31) (b) Kinematics. By considering the motion in the x direction: s x = (s B ) x + (v B ) x t s x = gh AB cos(α)t, and from the geometry x = l x = l cos α,

10 1 Motion of a single particle - Linear momentum, work and energy principle 1-10 thus, l cos α = gh AB t cos(α) l = gh AB t. (1.3) By considering the motion in the y direction s y = (s B ) y + (v B ) y t + 1 a yt = gh AB t sin(α) + 0.5( g)t, and from the geometry y = h BD + l y = h BD + gh AB t sin(α). Equating the geometrical quantity to the kinematic one (i.e. the vertical position of the skier with respect the chosen coordinate frame) we get thus, s y = y, 0 = h BD + gh AB t sin(α) 0.5gt. (1.33) Therefore the solution for t is: t = sin(α) gh AB ( sin(α) gh AB ) + h BD g g (c) Applying (1.34) and (1.3): t = s and l = 64. m.. (1.34)

11 1 Motion of a single particle - Linear momentum, work and energy principle Homework A m = 300 kg crate is dropped vertically onto a conveyor belt that is moving at v = 1. m/s. A motor maintains the belt s constant speed. The belt initially slides under the crate, with a friction coefficient µ = 0.4. After a short time, the crate is moving at the speed of the belt. During the period in which the crate is being accelerated, find the work done by the motor which drives the belt. (Note that the dynamic friction is not considered here.) Solution I. Choose the reference system Consider the reference system as shown in the figure of the assignment. Free body diagram for the belt and the crate. Since the gravity is balanced by the normal force, the only remaining, the only forces that remain are the two Newton-pair friction forces (F x ) - of the belt on the crate, and of the crate on the belt: F x = µmg. The key insight is to realize that only the belt matters here, because it is the belt s motor that we re interested in. We need to calculate the work exerted on the belt. We know the force, and so we need the distance the belt travels. Since we know how fast the belt is traveling, if we find how long the block takes to move with the belt, we can find how much it traveled. The initial velocity of the block is v 0 = 0 m/s, while the final velocity is v f = 1. m/s. The acceleration of the block is a = F x /m = µg. The time to reach the final velocity is v f = v 0 + at t = v f a = v f µg.

12 1 Motion of a single particle - Linear momentum, work and energy principle 1-1 Given this, the distance the belt traveled is d = v f t = v f µg. Finally the work done by the motor is W = F x d = µmg v f µg = mv f = 43J. (1.35)

13 1 Motion of a single particle - Linear momentum, work and energy principle Homework A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball s motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, the magnitude of the constraint force on the ball is N top and N bot, respectively. Find the difference N bot N top. (Find an expression in terms of m.) Solution I. Choose the reference system Normal-tangential (n-t) coordinates are attached to, and move with, the ball. Free-body diagram at the top and the bottom: Let v top and v bot be the velocities of the ball at the top and bottom of the circle respectively. At the top of the circle, applying the linear momentum principle in the normal direction: P n,top = F n,top mv top L = N top + mg. (1.36) Using the same consideration at the bottom of the circle: P n,bot = F n,bot mv bot L = N bot mg. (1.37) Therefore we get N bot N top = m L (v bot v top) + mg. (1.38) Since all forces are either potential (G = mg) or do no work (N bot, N top ), the system is conservative, thus, we can use conservation of energy principle to relate v top and v bot T top + V top = T bot 1 mv top + mgl = 1 mv bot, (1.39) 4gL = v bot v top. Substituting (1.40) into (1.38) (1.40) N bot N top = m 4gL + mg = 6mg. (1.41) L

14 1 Motion of a single particle - Linear momentum, work and energy principle Homework The kg collar is released from rest at A and travels along the smooth vertical guide. Determine the speed of the collar when it reaches position B. Find also the normal force exerted on the collar at this position. The spring has an unstretched length of 00 mm. Solution I. Choose the reference system Normal-tangential (n-t) coordinates are attached to, and move with, the collar. Free-body diagram at point A: Since all forces are either potential (F sp, mg) or do no work (N), the system is conservative, thus, we can use conservation of energy principle. With reference to the datum set in Fig.(a), the gravitational potential energy of the collar at positions A and B are (V g ) A = mgh A, (V g ) B = mgh B. (1.4) When the collar is at positions A and B, the spring stretches s A = l A l 0 and s B = l B l 0. Thus, the elastic potential energy of the spring when the collar is at these two positions is (V e ) A = 1 ks A, (V e ) B = 1 ks B. (1.43) Conservation of energy: T A + (V g ) A + (V e ) A = T B + (V g ) B + (V e ) B,

15 1 Motion of a single particle - Linear momentum, work and energy principle mv A + mgh A + 1 ks A = 1 mv B + mgh B + 1 ks B. (1.44) Since h A = 0 m and v A = 0 m/s: k(s v B = A s B ) gh B. (1.45) m Substituting s A = m, s B = m and h B = 0.6 m into (1.45): v B = 5.17 m/s. Free-body diagram at point B: Applying linear momentum principle in the normal direction at point B. P n = F n ma n = mg + F sp sin(θ) N B N B = mg + F sp sin(θ) ma n = mg + F sp sin(θ) m v B r Since the F sp = k(l B l 0 ) = ks B, substituting the numbers we get: (1.46) N B = 08 N. Note: the negative sign indicates that N B acts in the opposite direction to that shown in the free-body diagram.

16 1 Motion of a single particle - Linear momentum, work and energy principle Homework The man on the bicycle attempts to coast around the ellipsoidal loop without falling off the track. Determine the speed he must maintain at A just before entering the loop in order to perform the stunt. The bicycle and the man have a total mass of 85 kg and a center of mass at G. Neglect the mass of the wheels. Hints: If the path is defined as y = f(x), the radius of curvature at the point where the particle is located can be obtained from r = [1+(dy/dx) ] 3/ d y/dx Solution I. Choose the reference system Normal-tangential (n-t) coordinates are attached to, and move with, the cyclist. Now the free-body diagram is the following: Geometry. First calculate the radius of the curvature at point A and B. y = q p p x (1.47) dy dx = q p x(p x ) 1/ (1.48) d y dx = q p [x (p x ) 3/ + (p x ) 1/ ] (1.49) The slope angle at point B is given by: tan θ = dy dx = 0 θ = 0, (1.50) x=0

17 1 Motion of a single particle - Linear momentum, work and energy principle 1-17 and the radius of curvature at point B is r = [1 + (dy/dx) ] 3/ d y/dx =.5m. (1.51) Since the center of mass (h G ) for the cyclist is 1. m off the track, the radius of curvature for the cyclist is: r b = r h G = 1.05 m. Apply now the linear momentum principle in the normal direction at point B. In order for the cyclist to pass point B without falling off the track, it is required that N B 0, that is P n = F n ma n = mg + N B v B r b = g v B = gr b. (Note that N B = 0.) Setting the origin of the reference frame at the center of mass of the cyclist before he enters the track, when he is at point B, h with respect to his initial position (point A) is: h = y B y A h G. Since all forces are either potential (G = mg) or do no work (N B ), the system is conservative, thus, we can use conservation of energy principle. T A + V A = T B + V B 1 mv A = 1 mv B + mg h (1.5) v A = v B + g h v A = gr b + g h = 10.96m/s (1.53)

18 1 Motion of a single particle - Linear momentum, work and energy principle Homework The m = 0.5 kg ball of negligible size is fired up the smooth vertical circular track using the spring plunger. The plunger keeps the spring compressed at s = 0.08 m when s = 0 m. Determine how far it must be pulled back and released so that the ball will begin to leave the track when θ = 135. Solution I. Choose the reference system Normal-tangential (n-t) coordinates are attached to, and move with, the ball. Free-Body diagram at point B: Applying linear momentum principle in the normal direction at point B: P n = F n m v B r = mg cos(180 θ) + N B v B = rg cos(180 θ). (1.54) (Note that the limiting case is when N B = 0.) Since all forces are either potential (F sp, mg) or do no work (N B ), the system is conservative, thus, we can use conservation of energy principle. Here, the ball is being displaced vertically by h B = r + r sin(θ 90 ) and the spring force is given by F sp ( s) = k( s + s). Applying the conservation of energy, and considering that the elastic energy of the spring is totally transfered to the ball T A + V A = T B + V B s 0 s 0 F sp d s = mgh B + 1 mv B, (1.55) k( s + s)d s = mgh B + 1 mv B 0.5ks + (k s)s mgh B 1 mv B = 0. (1.56)

19 1 Motion of a single particle - Linear momentum, work and energy principle 1-19 Substituting expressions for h B and (1.54) into (1.56) leads to s = k s + k ( s) + k[mg(r + r sin(θ 90 )) + (0.5mrg cos(180 θ))] k Substituting values into (1.57) gives s = m.. (1.57)