Rotational Kinematics Question?
824167
The radius is actually 2/3 of the altitude.  Try that
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
373075
Acceleration= 10 rad/s2\n\nAngular vel w= 10 x t rad/s\n\nAngular displacement d= 10 x t^2 / 2  :(t squared)\n\nThey say that angular displacement is 150 rad in 5 seconds, this means in an interval between t and (t+5).\nd= 10 t^2 /2 \nd2-d1= 10/2 (t+5)^2 - 10/2 t^2 =150 rad\nt^2 + 10t + 25 - t^2 = 30\nt= 0.5 sec\n\nSo the wheel totates for 0.5 seconds before they start to count.  \n\nHope its clear enough.
785816
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
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
730213
Let u, v be the rotated x, y, rotated counter-clockwise by an angle t.\n\nThen x = u*cos(t) + v*sin(t), and y = -u*sin(t) + v*cos(t).\n\nPlug these into the equation (abbreviating c=cos(t), s=sin(t))\n\n(uc+vs)^2 + 4(uc+vs)(vc-us) + 4(vc-us)^2 + (5*sqrt(5))(vc-us) + 5\n= u^2*c^2 + 2*uvcs + v^2*s^2 + 4*uvc^2 + 4*v^2*cs - 4*u^2*cs - 4*uv*s^2 + 4*v^2*c^2  - 8*uvcs + 4*u^2*s^2 + 5*sqrt(5)(vc-us) + 5\n= u^2(c^2-4cs+4s^2) + uv(2cs + 4c^2 - 4s^2 - 8cs) + v^2(s^2 + 4cs + 4c^2) + 5*sqrt(5)(vc-us) + 5\n\nso to eliminate the xy term, we must have 2cs + 4c^2 - 4s^2 - 8cs = 0, but 2cs = sin(2t) and c^2 - s^2 = cos(2t), so we have\n4cos(2t) = 3sin(2t), or tan(2t) = 4/3, so t = arctan(4/3)/2 = 0.4636 radians.\n\nThe resulting equation is 5v^2 - 5u + 10v + 5 = 0.
764582
1.) I assume that you're going from degrees to radians.\nRemember: 360 degrees = 2π radians\nTherefore: 1 degree = π/180 radians\n\n28 degrees = 28 * π/180 radians = 0.489 radians\n32 degrees = 32 * π/180 radians = 0.559 radians\n41 degrees = 41 * π/180 radians = 0.716 radians\n\n2.) 2π radians = 360 degree\nTherefore: 1 radian = 180/π degree\n\n7.46 radian = 7.46 * 180/π degrees = 427.4 degrees \n\n3.) 360 degree = 1 revolution\nTherefore: 1 degree = 1/360 revolution\n.84 degree = .84 * 1/360 revolution = 0.00233 revolutions
973163
Relative to a woman in the car:\n\n(a) the translational velocity is ZERO since the woman is moving at the same translational velocity as the car relative to the ground.\n\n(b) the magnitude of the translational acceleration of the center of the wheel is also ZERO because the center of the wheel is NOT moving and NOT rotating relative to the woman.\n\n(c) v, velocity of the point at the top of the tire is = 80 km/h because that is how fast the car is moving, thus the wheel must also be rotating at the same speed.\n\n(d) a, acceleration of the point at the top of the tire is given by:\n\na = v^2/r = (22.22)^2/(0.33) = 1,496.45 m/s^2\n\n(e) v, velocity of the point at the bottom of the tire is = -80 km/h because at the bottom of the tire, the tire is moving in the opposite direction\n\n(f) a, acceleration of the point at the bottom of the tire is the same as (d) a = v^2/r = 1,496.45 m/s^2\n\nNow relative to a hitch hiker on the ground:\n\n(g) v at the center of the wheel is = 80 km/h or 22.22 m/s, since the car and the center of the wheel (which is not moving relative to the car) are moving at the same speed relative to the ground.\n\n(h) a at the center of the wheel is = v^2/r = 1,496.45 m/s^2, since the center of the wheel is rotating about the bottom point of the tire as the tire rolls on the ground.\n\n(i) v at the top of the tire is v(top) = ωd, where ω is the angular velocity of the rotating tire and d is the diameter of the wheel. The reason we use diameter of the wheel is because the top of the tire is rotating abou the bottom of the tire, not the center.\n\nω = v/r = (22.22)/(0.33) = 67.33 rad/sec\nv(top) = ωd = (67.33)(0.66) = 44.44 m/s\n\n(j) a at the top of the tire is = v(top)^2/d\n    = (44.44)^2/(0.66) = 2,992.29 m/s^2  \n\n(k) v at the bottom of the tire is equal to ZERO because the bottom of the tire is not moving relativ to the hitch hiker or the ground.\n\n(l) a at the bottom of the tire is also equal to ZERO since the velocity is zero.
102434
Part (a): \n   Given three values (Mrock = 1.2 kg, Xrock = 5 cm, Xstand = 14 cm), and we need to solve for one unknown (Mstick).  To clarify, Mrock means mass of rock, and Xrock means position of the rock on the meter stick.\n   Since you're being tested on this concept, I'll assume you know the equation for torque: T = Force x Lever arm.  We will also use the concept of Static Equilibrium, which is a fancy term that means that nothing is moving.  Since this system is in static equilibrium, we can use the equation T(clockwise) = T(counterclockwise), using the stand as our pivot point (14 cm).\n   To find clockwise torque, we need to find the Lever arm from the center of mass to the pivot point.  The center of mass is located at 50 cm, and the pivot point is at 14 cm, so our lever arm is 50 cm - 14 cm = 36 cm.  T(clockwise) = Mstick * g * 36 cm  (It's okay to use units of cm in this problem because it cancels out in the end.)\n   If the rock is hung from a massless string, we can assume that the force on the meter stick from the string, which is the tension in the string, is equal to Mrock * g.\n   Counterclockwise torque: Lever arm = 14 cm - 5 cm = 9 cm.  Torque(counterclockwise) = Mrock * g * 9 cm (plug in numbers later to simplify calculation)\n   Torque(counterclockwise) = Torque(clockwise) ==>  Mrock * g * 9 cm = Mstick * g * 36 cm  ==> g's cancel, manipulate equation to get: Mstick = (9 cm / 36 cm) * 1.2 kg = 0.3 kg\n\nPart (b):\n   Since the stand is not causing the meter stick to accelerate upwards, it must be exerting a force upward equal to the force the meter stick exerts downward.  The total force from the meter stick is equal to (Mstick * g) + (Mrock * g) ===>  (Mstick + Mrock) * g ===>  1.5 kg * 9.8 m/s^2  ===> 14.7 N\n   I hope this helps (I'm a physics major...this should be a reliable answer)
701707
The Average Linear Speed is {total distance / total time taken} = 65/ 5.7 = 11.4 m/s. \n\nDivide this value by the radius of the wheel to get the ‘The Average Angular Speed’. \n\nThe Average Angular Speed is = 11.4 / 0.02 = 570 radian per second.\n\nSince the motion is uniform,\n\n The Average Angular Speed = {Initial Angular Speed + Final Angular Speed} / 2   \n\nBut the Initial Angular Speed is zero.\n\nTherefore, the Final Angular Speed   is twice the Average Angular Speed\n= 2 x 570 \n=1140 radian/ second. (The answer for the second question)\n\nSince the Initial Angular Speed is zero,\n\nThe Angular acceleration is, “Final Angular Speed/ total time”.\n= 1140/5.7 = 200 [radian / second] / second. The answer for the first question)\n\n\nThe number of revolutions is found by “total distance / (pi x diameter of the wheel)” = 65 / pi 0.04 = 517.25 revolutions.
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
941552
Call the mass M, moment of inertia I, velocity v, angular velocity w, drop height h, gravity g, and axle radius r.\n\nYou do this by conservation of energy.  call the energy zero when the yo yo first starts out.\n\nThe potential energy then is: PE = -Mgh\nThe translational kinetic energy is: KE = (M * v^2)/2\nThe rotational kinetic energy is: RE = (I * w^2)/2\n\nBy conservation of energy, the total energy (sum of these three) never changes but stays at the initial value of zero.\n\nThe conservation of energy equation is:\n\n0 = PE + KE + RE\n\nYou need one more relation, the connection between the velocity and the angular velocity.  These are related because the yoyo rolls down the string so:\n\n w = v/r\n\nYou still need to do the acceleration and time portions.  You cna take the derivative of the energy equation to get the time behavior.   After taking the derivative, remember that dh/dt  = v.
982223
You can't answer this without knowing the tape thickness.  Of course, you can solve it generally for any set of variables.  Call the initial radius (10") D, the hub radius (3") H, the tape thickness b, the rotation rate w (in rpm) and the instantaneous radius r.\n\nThe radius decrease by b after each revolution (call the revolutions x).  The rate of change of the radius with time t is:\n\ndr/dx = -b\n\nThe definition of rotation is: dx/dt = w\n\nBy the chain rule: dr/dt = dr/dx * dx/dt = -bw\n\nAs expected, the radius decreases at a constant rate.  The radius after 1 hour is:   D - bw * (1 hour)\n\nThe amount of tape unwound is just the difference in the area of the roll divided by the tape thickness:\n\nL = pi*(D^2 - (D - bw * (1 hour))^2)/b
759847
Fir the first question, you need to determine how many revolutions per second that is.\n\nSince one revolution is 2 * Pi radians, divide 4.60 radians per second by 2 * Pi to get the number of revolutions per second:\n\n4.60 / (2 * 3.14159) = 0.7321 revolutions.\n\nEach full revolution of the forward sprocket would pull a length of chain equal to the circumference.  Thus one complete revolution of the forward sprocket would pull a length of chain equal to (2 * Pi * radius) which is to say (2 * 3.14159 * 9) = 56.5486 centimeters of chain.\n\nThus, 0.7321 revolutions per second would pull (0.7321 * 56.5486) centimeters of chain per second, which is 41.3992 centimeters per second.\n\nReducing digits to account for significant figures, call it 41.40 centimeters per second is the linear velocity of the chain.\n\nCentripetal accelleration is velocity squared divided by radius, so (41.40 cm/s)^2 / 5.10cm:\n\n(41.40 cm/s)^2 / 5.10cm\n(1713.96 cm^2/s^2) / 5.10cm\n336.07 cm/s^2
755943
If you are assuming "uniformly accelerated rotation" then the following apply;  (please excuse the crude variables)\n\nw = dθ / dt\na = dw / dt = d^2 θ / dt^2\n\nwhere "w" stands for omega <-- angular velocity\nand "a" stands for alpha <--  angular acceleration.\n\nIf you have had any calculus you can take the first derivitive of your θ expression to get the angular velocity, and then the second derivitive to get the angular acceleration.\n\nsince we have a a simple polynomial the following says\n\nd/dx [x^n] = nx^(n-1)  , however in our case we will be deriving t\n\nTo yeild       w = 8t + 6t^2   , for angular velocity\nand              a = 8 + 12t      , for angular acceleration\n\n\n(a) if you plug in 0 for t then you get θ = 2radians\n(b) its angular velocity is zero when t = 0\n(c) w = 8(4) + 6(4^2) = 128 radians/second\n(d) a = 8 + 12(2) = 32radians/second^2\n(e)  I don't believe its angular acceleratin is constant.  Because if you do some testing with plugging in times for acceleratin and velocity they don't quite add up...   \n\nBut that could be because I am totally wrong.  Anyway that is my stab at it.  Hope it helps
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)