a solid cylinder of radius 10cm and mas 12kg starts from rest and rolls without slippig a distance L=6m dwon a roof that is inclined at angle 30(degrees). what is teh angular speed of teh cylinder about its center as it leaves teh roof?
206629
the acceleration of the center of mass as the body rolls an an inclined plane is given by,\n A=g*sin(angle)/[1+I/MR^2]        (for proof see any dynamics boook)\n I=0.5MR^2\n  =0.06\ni.e.  A=3.27 \ntherefore the speed of the CM after travelling 6m is\n v=sqrt(2AL)\nv=6.26\nangular velcity W=v/R\n                W=6.26/0.1\n                W=62.6 rad/s
363498
This problem requires you to use the conservation of angular momentum, and then calculate the change in moment of inertia. Basically, moment of inertia is the rotational equivalent of mass. Moment of Inertia is defined as:\n\nVolume in tegral of ( r² dm )\n\n(I have no good way of writing this - check the Wikipedia article below)\n\nNext, the Conservation of Angular Momentum can be written as: \n\nI * w1 = I * w2\n\n(a) To solve this problem, first, note that both the carosel and the man have the same rotational speed of .15 rev/s. Second, assume the man is a point mass. Then, the moment of inertia for the system can be calculated as:\n\nI-carosel = (1/2) * M * R² (defined for a cylinder)\n\nI-man = m * r1² (initial)\n\nI-man = m * r2² (final)\n\nThe net moment of inertia for the initial sysem is simply:\n\nI-initial = [(1/2) * M * R²] + [m * r1²]\n\nI-initial = [(1/2)(25 kg)(2 m)²]+[80 kg * (2 m)²]\n\nI-initial = 370 kg-m²\n\nThe net moment of inertia for the final sysem is simply:\n\nI-final = [(1/2) * M * R²] + [m * r2²]\n\nI-final = [(1/2)(25 kg)(2 m)²]+[80 kg * (.7 m)²]\n\nI-final = 89.2 kg-m²\n\nUsing the conservation of angular momentum:\n\nI-initial * w1 = I-final * w2\n\nRe-arranging, we can solve for w2:\n\nw2 = (I-initial / I-final) * w1\n\nw2 = (370 / 89.2) * (.15 rev/s * 2 * pi rad/rev)\n\nw2 = 3.91 rad/s\n\n(b) The kinetic energy of a rotating body is given as:\n\nKE = (1/2)*I*w²\n\nThus, the change in kinetic energy is:\n\nKE = (1/2) ( I-final * w2² - I-initial * w1²)\n\nKE = (1/2) ( 89.2 kg-m² * (3.91)² - 370 * (.15 rev/s * 2 * pi rad/rev)²)\n\nKE = 517.3 J\n\nThat's it! :)
373547
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
480315
You need to look at the normal acceleration versus the acceleration due to gravity at that angle.\n\nThe normal acceleration = v^2/R (where R is the radius and v is the velocity) and if this is greater than the component of gravity pointed towards the center of circle then airborne. So as it slides down the circle the velocity, v, is increasing, at the same time the component of gravity gets smaller.\n\nAt 0 degrees  (on top) v is zero but all the gravity is pointed towards the center of the circle. At 90 degrees the velocity is equal to square root of (2 * g * R) but the gravity component passing through the center of circle is 0. So somewhere between 0 and 90 degrees it takes off.\n\nI will let you do the math, but I would check 48 degrees as your instructor implied. (wink wink), use conservation of energy to find v.
405901
h = 4tanθ\nVh = dh/dt = (dh/dθ)(dθ/dt)\ndh/dθ = 4/cos^2θ\ncos^2θ = 4^2/(4^2 + 19^2)\nVh = (16 + 361)/4)(5/361)\nVh = (5/4)(16/361 + 1)\nVh = 1.3054 km/sec
384767
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
165307
When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.\neverything in usual notation,\nacceleration A=R*a (a=angular acceleration)\nNewton's second law implies'\nMA=F-f         (resultant force)\nalso for torques,\nIa=FR+fR         (for the axis passing through the centre)\nFRom the two equatins,\n2F=mA+Ia/R\ni.e.  2F=maR+Ia/r\n   a=2F/(mR+I/R)\nI=0.5MR^2=0.05\nplugging hte values,\na=16 \ntherefore, A=aR\n            =16*0.1\n           A=1.6
542139
OK, let us begin\n\n1. 1 reverse is 2Pi radians - 6.28 radians, so angular speed in radians will be \n     W = 6.28(rad/rev)x1750(rev/sec) = 10990 (rad/sec)\n2. The linear speed at a point 1.38 cm from center will be the angular speed in radians multiplied by the distanse from the center:\n     V = W x r = 0.0138(m)x10990(rad/sec) = 151.662 m/s\nHere we expressed the distance 1.38 cm as 0.0138m\n3. The radial acceleration on the point of the rim. The distance from the center to the rim R is 3.53 cm or 0.0353 m, \n  the radial acceleration is Arad = W V(R) = W (W x R) =\n   = W^2 x R = 10990x10990x0.0353 m/s^2 = 4263537.5 m/s^2\n\n To check our result, let us find the linear velocity of a point on the rim   V(R) = W R = 10990x0.0353 m/s =387.947 m/s\nCheck Arad = W V = 387.947m/s x 10990 1/s = 4263537.5m/s^2 \n 4. Linear velocity at the rim is 387.947 m/s,\na point on a rim will cover a distance in 2.75 sec as following:\nD = V T = 2.75 (s)x 387.947 m/s = 1066.854 m\n\nSorry for possible grammar mistakes. I hurried, hope your lesson has not finished yet.
333168
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
444510
The 2nd and 3rd answers match one of the choices provided.\nThe first one is a bit off.\n\nThe component of the gravitational acceleration vector acting along the line of the string is 9.8*sin(x) m/sec^2 where x is the angle between the string and the horizontal. The centripetal acceleration is given by v^2/r.\n  \n1) a =2.7 m/sec is closest\n\n At a 40 degree angle above the horizontal the gravitational acceleration component reducing the tension in the string due to the centripetal acceleration is: 9.8xsin(40) = 6.3 m/sec^2\n\nThe centripetal acceleration = v^2/r = v^2/1.4m\n\nThe net acceleration (tension) = 3N/0.3kg = 10 m/sec^2\n\nv^2/1.4 = 10-6.3\nv^2 = 5.18\nv = 2.3 m/sec\n\n2) b. 11 N\n\nThe centripetal acceleration is: \n\nv^2/r = 8^2/2 = 32 m/sec^2\n\nThe gravitational acceleration component, at the top, which reduces the tension is 9.8 m/sec \n\nThe net acceleration is: 22.2 m/sec^2\n\nThe tension (net force) = ma = 0.5 x 22.2 = 11 N (rounded off)\n\n\n3) b. 4\n\nlet v = the wheels angular speed at the start of the 4 sec\n\nd = 80rad = v*t + 0.5 at^2 \n\nt=4sec; a = 4rad/sec^2\n\n80 = 4v+0.5*4*(4^2) = 4v+32\n\n20=v+4\n\nv=16\n\nIf the wheel starts from rest:\n\nv=a*t =16 = 4t\n\nt = 4sec
531381
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
528992
^ Parasite alert LOL! ^\n\n30 s for C to move from A to B, maybe? I'll assume that. Also, x seems to be a length, not an angle. Maybe a) is asking for the rate of change of z (called z')? We know the angular rate of C about O (y') is pi/30 (answer to b)), and triangle CAO is isosceles, so z'=-0.5*pi/30. \nFor c), circumferential velocity is pi*r/30. Angle CAO is, conveniently, 45 deg, so x' = sqrt(0.5)*circumferential velocity.\nFor d), the area of triangle OAC is .5*AO*altitude, and altitude=r*sin(y). This is a maximum when y (not x) =pi/2.\nSorry, I know you wanted  no answers, but I couldn't see how/where to stop short of most of the answers.
502974
Well,\n\n1 C. from F=ma, since F is non zero, then a has to be non zero.\n\n2. vector addition.  work it out.\n\n3. A  0.4\n\nWork out the decelerration.  This force acts on the crate,\nsince the crate does not slip, its balanced by the\nuN force of kinetic friction.\n\nso uN = ma, u = ma/N = ma/mg = a/g.\na is the decceleration\ndistance travelled d = vt + 1/2at^2,\n and t = v/a\nwhere v = 15 and d = 28.7, so two equations in two unknows.\n\n4. vector addition 27n to the right
361143
Tell the other person to stop and you run at an angular speed that's the sum: (1.5 + 3.2)*10^-3 rad/s. How long before you cover the whole 2*pi*radians of the lake?
527006
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
314578
Homework?\nOK, only a hint. Consider a cylinder made of the circle and with a height dh. What is the volume of this cylinder? \n\nNow displace the cylinder along the path over dh, etc.
373628
A right cylinder is a cylinder in which a line going straight through would be parallel with an end.\n\nYour problem would be V=pi*r^2*h\nso dV/dt= pi*(r^2* (dh/dt) +h*2r*dr/dt)\n\nSubstituting in the values, you get 0=pi*(5^2 *(dh/dt)+20*10*0.5)\nThen, you'd get 0=pi* (25*(dh/dt)+100)\nDividing by pi on both sides and subtracting 100 on both sides, you get -100=25*(dh/dt)\nSo that means the answer is -4 cm/min