Exam1-soln-TTh

Exam1-soln-TTh - Problem 1. This problem can be solved in...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 1. This problem can be solved in two ways. One way is to use a = vdv/dx. The value for vo follows directly from the asymptotic value of x. The second way is to use the direct definition of acceleration, viz., a = dv/dt. Once the velocity is known, we solve for position by integrating dx/dt = v. ... ...... ... ... . ... ....... ... . ... . ... .. ... ... ...... .... .. ... ..... ..... ... ... ... ..... ...... ..... ... ..... . ... ... ... ... ... ... ... .... . . ... ... ... ... ... .. ... ... ... ... ... .. ... ... ... ... ... ..... ... ..... . ... ... ... ... ... ... ... .... ... ... . ... ... ... ... ... ... ... ... ... .... . ... ... ... ... ... ... ... .... . ... ... ... ... ... ... ... .... ... ... . ... ... ... ... ... ... ... ... ... .... . ... ... ... ... ........................ ... ... ... .... . . .. . ... ....................................................... . ... .. . ... ................... .............. ... ... ... .................................................... .. . . . .. . ... ........................................ ... .... . . ... ..................................... . . ... .... ............ ........ ... ... ... .... ...................... ... . ... .. . . ... . . .. . ...... ......................................... ........ .... . ...................... .... . ...... ..................................... ................... ..... .................. ........ .. . . ............ ............................ . . . .. . ................ ........ . ....... . ....... ...... .. .... .......... . . ... .. . ............. .......... ............... ........................... . ............. ............ ................. ............................ . . ..... . . . . . . . .. ...... ... . . ........ ................ . ............... . . ................. .... ............................. ............................... ......................... .................. ................................................... . . . . ................................ . ....................................................... . . . ... . . ................................ ........................................... .................................. ........................................................ . . . . . . . . . ................................ . ....................................................... . . ................................ ................................ . ....................................................... . . . . . .... . . . . . . . . . . . . . .. . . . ................................ . ................................. . .. . . . ..... .. .. .. .. ...... . ... ... .. . . . . . . ... ... ... ..... .......................................................................................... ..... ... ... . . . . . .. . . ....................................................................................................................................................................................................................................................................................................................... ...... .. ...................... ................................................................................................................................................................................................................................................................................................................. . . .. .. .. .... ..... .. ............ . . . . .. .. .. .. . . ................................ ..... ......... . . . ... ... ... .. .. ............................................................................................................................................................... . . . . . .. ...... .. . ..... .. ... ..... ........ .. ...... ....... .. ........... . . .. .... .. ... ... .. .. .. .. . ... . .. . .. . .. ..... .................................... . ... . .... . . .... . ... . ... ....... . ...... . ............................................................................................................................................................................................................................................................................................................................................................ ... . .. ... ... .. ... .... ... .. . ... ..... ..... . ...... .. . .... .. . .. ... . .. . . . . . ... ....... . .. .... ... ......... ... ... ... . . . ... ............ . . ... .... .... .... ................ .. .......... ...... . ........... .. ... ..... ........ .... .. ... .. ... .. . .......................................................................................................................................................................................................................................................................................................................................................... . ......................................................................................................................................................................................................................................................................................................................................................... ....... ...... . ..... ............ ....................... ............. ... ............................................................................................................................. ..................... . .. . . . . . . . . . . . . .. . . . . . .. ... . . . . . . . . .. . . . . . . . . . ............................................ ... .............................................. .............................................. . ........................................... . . . .. . .. . . A x Solution using a = vdv/dx: For this approach, we will determine the slider's velocity as a function of position. Hence, a=v Dividing through by v, we have dv = - dx = v = vo + -x dv dx = vdv = a dx = - vdx where we have made use of the fact that v = vo and x = when t = 0. We are given that, as t , the position x 3 . Since position has a fixed value in this limit, clearly the slider 2 comes to rest and v 0. Thus, we can say that 0 = vo - Therefore, solving for vo yields vo = 2 2 Solution using a = dv/dt: For this approach, we will have to determine the slider's position as a function of time. Hence, from the definition of acceleration, we have a= Rearranging terms, we have dv dt =- v = n v = n vo - t dv dt = v dv = a dt = - dt where we have made use of the fact that v = vo when t = 0. Thus, the slider's velocity as a function of time is v(t) = vo e-t/ 1 We can now determine the slider's position by using the definition of velocity, viz., v= dx dt = dx = vo e-t/ dt = -vo d e-t/ dt Integrating and using the fact that x(0) = , x(t) = - vo e-t/ - 1 Therefore, the slider's position as a function of time is x(t) = ( + vo ) - vo e-t/ Now, we know that when t , the slider's position asymptotes to t 3 2 . So, lim x(t) = + vo = 3 2 = vo = 1 2 Therefore, solving for vo yields vo = 2 2 Problem 2. The easiest way to solve is to first compute the ball's motion as seen by a fixed observer and to then compute the cart's location for its given velocity. Equating the two values of horizontal position provides the information need to solve for the angle . y ................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................. ... .............................................. . .. .. .. . . .. . .. ... ... . .. . ......... .......... . . .. ... . ... .. . .. .. . .. .. . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. .. .. .. .. .. .. .. ... .. ... .. . ... .. .... .. .. .... .... .. . .... . .. .. ... ... .. . . ... . ... . . ......... ......... ... . ... . . .. . .. . .. .. .. . .............................................. .............................................. y O .............................................................................................x........ ...................................................................................A......... ................f...........r..........................................................r...........f............x . . ..... ....................................................................................................................................................................................................................................................................................................... ............ .......... . .. .... .......... . ......................................................................................... . . . .. .... ... ... ... . . . ... ... ... ............................................................................................................................................................................................................................................................................................................................................................................................................................... . . .. .... . ......................................................................................................................................................................................................................................................................................................................................................................... ..... .............. .... .... ................................. ............................................................................................................................................................................................................................................................... . ............. .. . . ...... ..... . . . . ............ ............ . ...... ..... . . . . ............ . . .... t Solution. First, note that the absolute velocity of the ball when it is launched is vB = xB (0) i + yB (0) j = (U - V sin ) i + V cos j Integrating once and using the initial conditions xB (0) = yB (0) = 0, the absolute position of the ball is 1 xB (t) = (U - V sin )t and yB (t) = V cos t - gt2 2 When the ball lands on the cart at time tf , we know that yB (tf ) = 0, wherefore 1 V cos = gtf 2 = tf = 2V cos g The position of the back of the cart is given by t xA (t) = 0 1 (U - at )dt = Ut - at2 2 Hence, the ball's position relative to the translating cart is xB/A (t) = (U - V sin )t - (U t - at2 /2) = (at/2 - V sin )t We seek the solution for which xB/A (t) = 0. This is true when the ball is launched at t = 0 and at t = tf when it lands, where tf is given by tf = 2V sin a We now have two equations for tf , one from above corresponding to the condition yB (tf ) = 0 and the other for xB/A (tf ) = 0. Equating the two expressions for tf yields tf = Therefore, the angle is = tan-1 a g 2V sin 2V cos = a g 3 Problem 3. Note that, with defined as indicated in the figure, positive velocities and forces are in the clockwise direction. z g z R R z ...................................................................................................................................................................... ........................................................................................................................................................................... .. . .. ... .. .. . . ... .. ..... ..... .. ....................................................... ...... ................................................................................ ........................................................ .. . ................................................................................................................................................................................... ................................................................................................................................................................................................ ................................................................................... ...... ...... .................................. .. .. . . . . . . . .................................................................................................................................................................................... . . .. . .. . . . . . . . . . . . . .. ........................................................................................................................................................................................................................................................................... . . .. . . . . . . . . . . . . . . . . . . ... .. ........................................................................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................................................................... .................................................................... . ............ ..... ............................ .. ................................. . . . . . . . . . . ... . . ................................................................................................................................................................................. ........................................................................................................................................ . .................................................................. . .................................................................................. .......................................... ....... ..................... ....... ....... ..................... ..................... .. . . . ............. . .. . ........................................................................................................................................................................................................................................................................................................................................................................................................ .................................................................................................................................................................................................................................................................................................................................................... ................................................................... .... . .... .... .... ..... .... .... .... .... .... .... .... ..... ..... ......... .... ...... . .. .. . . .. .. .. . . .. . . (a) Newton's Second Law tells us that Fr = m r - r2 and F = m r + 2r Up to the point of separation, the ball moves on the cylindrical surface, wherefore r = R and r = r = 0. Therefore, Fr = -mR2 and F = mR There are two forces acting on the sphere, viz., the ball's weight, mg, and the normal reaction force, N . As shown below, mg acts vertically downward and N acts radially outward. N mg z mg sin mg cos Thus, the equations governing the motion up to the separation point are N - mg cos = -mR2 and mg sin = mR (b) Now, multiplying the circumferential equation through by we find mg sin = mR Integrating once yields 1 -g cos = R2 + constant 2 But, = V /R when the sphere is at the top of the cylinder where = 0 Thus 1 V2 -g = R 2 + constant 2 R Solving for 2 , we find V 2 2g 2 = 2 + (1 - cos ) R R 4 = 1 1V2 - g cos = R2 - -g 2 2 R = g 1 d 2 d (- cos ) = R dt 2 dt (c) Substituting the result for 2 from Part (b) into the radial equation of motion, we have N - mg cos = -mR Thus, solving for N yields V2 N = -m + mg(3 cos - 2) R (d) When the normal reaction force vanishes, the sphere has reached the angle, s , at which it separates from the cylinder. Setting N = 0 in the equation derived in Part (c), there follows mg(3 cos - 2) = m Thus, the separation angle follows from 3 cos s = 2 + We are given s = cos-1 ( 5 ). Hence, 6 3 Thus, simplifying yields V2 1 = gR 2 5 V2 =2+ 6 gR = 5 V2 =2+ 2 gR V2 gR V2 R V 2 2g V2 + (1 - cos ) = -m - 2mg(1 - cos ) R2 R R 5 ...
View Full Document

Ask a homework question - tutors are online