Physics, Moment of Inertia Please help Today!?
829645
This is half a question! Would you be able to understand this question without more info? Are answerers supposed to be clairvoyant? But... read on.\nHow was the calculated MI calculated? Experimentally?\nYou show a 3.6% error. This could be considered reasonable experimental/measurement error. Could you have ignored (perhaps you were told to ignore) some contributors to error, such as: mass of a connecting string; error in pulley diameter (exactly what length of string per revolution can be difficult to determine because of string thickness); error in effective radius of mass of pulley (if it's the pulley MI you're dealing with) or wheel. Add more information to this Q via the "Additional Details" option and you may get a better answer.
742316
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.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
757946
The moment of inertia of the turntable plus the child at radias R is I + m*R^2. Setting up the equation of conservation of momentum\nI*(1 rev/T) = (I + m*R^2)*w Solve for w.
984692
The moment of inertia of a solid cylinder is 1/2 mr ^2..\nTorque is the product of moment of inertia and angular acceleration.  It can also be expressed as the product of the force applied to the edge of the disk and the radius.  Set the two products equal to each other and solve for m.\n\nT = I α = r F\n\n1/2 m r ^2 α = r F\n\nm = (r F) / (1/2 r ^2 α)  = 2 F / r α\nm = 2 (40) / 0.16 (90)\n\nm = 5.56 kg
906380
First problem: You have assumed all the mass of the meter stick is concentrated at the lower end; obviously, it is uniformly distributed along the length of the stick.  Assume that the mass is concentrated at the center; then, the length of the oscillator will be 0.865 meters, and the period should correspond.  There are some complications involving moment of inertia which I am ignoring.\nSecond problem: You may be solving for the wrong thing.  It looks like you are solving for the frequency of oscillation, rather than the maximum speed of motion.  Some careful work with units will help show you what is going on.
746380
Torque = change in angular momentum over change in time\n\nT=Iw/t      T=mrrw/t     T=1.1(.09)(93.6)/1.7  T=5.45 N/m
793054
The moment of inertia of the disk is 0.5*m*r^2 (ref.). (In kg-m-s units, m=.017 and r=.06.) The average angular acceleration A=23/0.86 rad/s^2. The torque T is the product of moment of inertia and angular acceleration. T=0.5*m*r^2*23/0.86 N-m.
1092161
all you are doing is using electricity to create angular moment, then hoping that the resulting angular moment will produce more electricity than required to create it, while at the same time overcoming the drag coefficient. Won't work.  There is nothing magic about angular momentum. It still involves velocity and mass. \nYou might look at taking an engineering dynamics class, usually sophomore year, or pick up a calculus based physics class, also sophomore year college level.\n\nSorry for the bad news
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
811117
1)))  inertia would be best described as\n       resistance to acceleration\n       a property of matter\n2)))zero\n\n3)))Yes\n\n4)))a))please specify the direction in which you want to calculate force\n    b))please specify whether the body was going raght or left\n\n5)))We experience the pseudoforce
707580
The moment of inertia is the resistance to change in rotational motion analogous to the mass for linear motion. If you think about the definition of energy = 1/2mv^2,   using w is angular velocity  we want to define I so that 1/2 Iw^2 is also the energy. But v=rw so mv^2=mr^2w^2 so if we define a variable I=mr^2, then we have the desired result, the energy of an object rotating at a angular velocity w will be 1/2 Iw^2. You can do calculations with out using I but it is more tedious
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
950937
You're in luck.  I teach physics and I know exactly how to answer this question (it's based on the work-energy principle).  It's pretty simple, really.  But it would be best if we discussed it by messenger.  Yahoo Messenger me, my ID is\nfortitudinousskeptic\n\n- Kevin
922680
The moment of inertia of a sphere is:\n\n(2/5)* M * R^2\n\nFor a hollow sphere (spherical shell) you can subtract the hollow part from the sphere part, but since the M (mass) term will change based on diameter it is easier to substitute:\n\nM = p * 4/3 * PI * R^3     (p = density)\n\nSo.. in the end you get  I = 0.5333 p ( R2^5 - R1^5 ) \n\nWhere R2 = Outer Radius, R1 = Inner Radius