* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download exam 2
Survey
Document related concepts
Derivations of the Lorentz transformations wikipedia , lookup
Center of mass wikipedia , lookup
Hunting oscillation wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Coriolis force wikipedia , lookup
Centrifugal force wikipedia , lookup
Classical mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Seismometer wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Transcript
Physics 218 Honors Exam 2 Please work 5 of the 6 problems, and cross out the one not to be graded. 1. A sawhorse is constructed of 5 identical boards as shown. Each board has a length of 1 m. The angle between each pair of legs is 60o. How high above the base is the center of mass? If the sawhorse were set up on an inclined plane as shown, what is the maximum tilt angle θ for which the sawhorse would not topple? Solution: By symmetry, the center of mass lies along the vertical midplane of the assembly. Each leg makes an angle of 60o with respect to the base. The cm of each leg L sin 60° therefore has a height hleg = = 0.433m 2 The cm of the top board is htop = L sin 60° = 0.866m The cm of the total assembly is then 4hleg + htop hcm = = 0.52m 5 The sawhorse can be tilted until its center of mass is directly above a pair of legs. Beyond that point, the horse would topple. hcm tan α = = 1.04 Let α be the angle between the cm and the base. L cos 60° α = 46° The angle through which the sawhorse can be tilted is then 90° − a = 44° . 2. A bicycle has 1 m diameter wheels. It is traveling along a straight road, with a linear velocity of 5 m/s and a linear acceleration of 1 m/s2. Calculate the vector velocity and vector linear acceleration of dots on the wheel at points A, B, C, D, as seen by a stationary observer on the side of the road. Give magnitude and direction in each case in a clear manner. Solution: The linear velocity is a vector sum of the translational velocity and the rotational velocity. The acceleration is a vector sum of the translational acceleration, the tangential acceleration, and the centripetal acceleration. The centripetal acceleration is a c = v 2 / R = 50 m/s 2 r ) v = (+5 − 5) x = 0 A: r ) ) ) a = (+1 − 1) x + 50 y = 50 y m/s 2 B: r ) ) v = +5 x − 5 y m/s r ) ) ) ) ) a = +1x -1 y-50 x = - 49 x -1 y m/s 2 C: r ) ) v = ( +5 + 5) x = 10 x m/s r ) ) ) ) a = ( + 1 + 1 )x -50 y = +2 x -50 y m/s 2 D: r ) ) v = +5 x + 5 y m/s r ) ) ) ) ) a = +1x + 1 y + 50 x = 51x + 1 y m/s 2 3. A uniform disk of mass 10 kg rolls without slipping up an inclined plane of angle o 20 . Its initial velocity along the incline is 10 m/s. How far along the incline does the disk roll before stopping? Please show a figure and neat work! Solution: ω Consider forces parallel to the slope. As gravity slows the linear motion of the disk, friction will act to slow the rotational motion as well; hence the friction force will point up the slope. Newton’s law for forces gives − mg sin 20° + F f = ma Newton’s law for torques gives − F f R = Iα 1 MR 2 2 Finally we require that there be no slipping: a = Rα. I= F f = Ma / 2 So 2 g sin 20° = 2.22 m/s 2 3 Now use the equation of motion along the slope: 1 x = v0 t − at 2 2 v = v 0 − at a= The disk comes to rest when v=0 → t = v0 / a = 4.54 s x = 22.7 m 4. A tetherball is hit so that it travels along a circular path around the pole as shown. The ball has mass 1 kg, the rope is massless and has length 2 m. The linear velocity of the ball is 2 m/s. What is its angular frequency, what is its moment of inertia about the pole, and what is the tension in the rope? (you can treat the tetherball as a point object.) Solution: We must first find out what is the angle θ between the rope and the pole. The circular orbit of the ball on a radius R = L sin θ requires a centripetal acceleration v2/R in the horizontal plane. The direction in which the ball is free to move is the circle of radius L about the pivot. We must use Newton’s law on the forces in this direction of equilibrium: v2 mg sin θ = m cosθ R Putting in the dependence of R on θ enables us to solve for R: v2 g ( R / L) = R R4 = ( R2 1 − 2 L v4 2 L − R2 g2 1/ 2 ) 2 2 4 v 4 − 1 + 1 + 4g L / v = 0.74 m 2 2 g2 R = 0.86 m, θ = 64o. R2 = v = 2.71 rad/s I = MR 2 = 0.93 kg m 2 R Applying Newton’s law to the vertical forces on the ball yields − mg + T cosθ = 0 T = 13.4 N ω= 5. Calculate how high above the table one must hit a cue ball of radius R in order that it roll without sliding as it leaves the cue. Assume the cue stick strikes the ball horizontally. Show your work and make a figure. Solution: Rolling without sliding requires that a = Rα. The cue exerts both a force F and a torque τ = F h. Newton’s law for force yields F = ma Newton's law for torque yields τ = Fh = Iα 2 The moment of inertia for a uniform sphere is I = mR 2 . 5 Putting it all together, 2 h= R 5 The ball must be stuck at a height of 7/5 R above the table surface. 6. A spherical professor (M = 100 kg) has climbed to the top of a 5 m long ladder (m = 20 kg, uniformly distributed along its length) that leans against a wall. The angle between ladder and floor is 60o. There is no friction with the wall, and a static friction coefficient of 0.55 with the floor. An angry student pulls horizontally on the ladder at its base. How hard must she pull in order that the ladder begins to slide? Solution: Horizontal forces: Vertical forces: Fw = force of wall on top of ladder, -Ff = friction of ladder on floor, F = force that student uses to pull ladder to the left. -Mg on professor, -mg on ladder, +F⊥ of floor on ladder. Newton’s law of forces: Fw − F f + F = 0 − Mg − mg + F⊥ = 0 We obtain F⊥ = 1176 N. The maximum friction force available is F f = µF⊥ = 647 N Newton’s law of torques (take pivot at the bottom of ladder): L Fw L cos 60° − MgL sin 60° − mg sin 60° = 0 2 We obtain Fw = 622 N. The additional force needed if the ladder is to slide is F = 25 N.