Determine the torque applied to a cylinder?
Draw a FBD of the cylinder. You should get the following equation: Sum of the moments about the center of the cylinder = I*alpha = M - T*d/2 *** T is the tension in the rope and M is the torque applied *** Solve for M M = I*alpha + T*d/2 To find the tension in the rope, do a FBD of the mass. You should get the following equation: Sum of the vertical forces on the mass (up is positive) = m1*a = T - m1*g T = m1*(a+g) The moment of inertia for a cylinder about the axis running through the center of the circular cross section is... I = 1/2*m2*r^2 = 1/2*m2*(d/2)^2 = 1/8*m2*d^2 The angular accelertaion can be found through the linear/angular relationship alpha = a/r = 2*a/d Sub all this into the M equation M = 1/8*m2*d^2*2*a/d + 1/2*m1*(a+g)*d M = 1/4*m2*a*d + 1/2*m1*(a+g)*d Now plug in the numbers M = 1/4 * 22 kg * 2 m/s^2 * 6 m + 1/2 * 16 kg * 11.81 m/s^2 * 6 m = 630 N*m (2 sig figs) Hope the explanation helped.
The lug nuts on a car wheel require tightening to a torque of 90 N ∙ m. If a 30-cm long wrench is used, what is the magnitude of the minimum force required using the wrench?
First it is necessary to convert everything to the same units (30cm needs to be converted to meters).30cm * (1m/100cm) = .3mTo find force you need to use the torque equation to solve for force.T = F * dSo,F = T/dWhere,T = torqueF = forced = lever armIf you plug 90Nm and .3m into the above equation you will find the answer to be 300N.
Torque, moment of inertia and finding angular speed?
This is a 2 step problem. 1) Find the angular acceleration, alpha 2) Find the angular speed First the acceleration (1): torque = I*alpha, where I = moment of inertia alpha = angular acceleration in rad/s^2 Solve: alpha = torque/I = 27.1N*m/.128kg*m^2 = 211.7 rad/s^2 Now the speed. Do you remember the equation for linear motion, v^2 = vi^2 + 2*a*s ?? This is the time-independent motion equation and is derived from conservation of energy. We use the angular form: omega^2 = omega_initial^2 + 2*alpha*theta, where omega = angular frequency in rad/s omega_initial = initial angular freq alpha = angular acceleration from above theta = rotations in radians I'm going to assume that omega_initial = 0 Solving: omega = [2*alpha*theta]^1/2 omega = [2*211.7 rad/s^2*15.7rev*(2*pi rad/rev)]^1/2 omega = 204 rad/s = 204 rad/s * 2*pi rev/rad = 1280 rev/s Note that we had to convert revolutions to radians to use our formula. There are 2*pi radians in one revolution.
How can I calculate the force needed to turn a generator?
You need the power output, the efficiency of the generator, and then you say what force... hm.In what way are you applying this force? Say you are turning a crank handle connected to the generator. The force it requires depends not only on the power, but on the length of the crank. You'd be pushing it with a given force in newtons sufficient to drive the generator with a certain amount of power at a given speed. So the length of the crank must be taken into account.Perhaps you meant to ask how much power you need to turn the generator, in horsepower or watts. That is a more meaningful question.In that case, the power required is simply the power output divided by the efficiency of the generator including any gears or pulleys. That gets you an answer in watts. For horsepower, divide that number by 745.7 as there are 745.7W in one horsepower.
Why do you not get as much torque when force is applied closer to the axis of rotation?
This is a matter of leverage, which, mathematically, is a matter of angular momentum.I am about to get a bit physicsy, so bear with me. The main thing to know is that when a wheel or gear spins, the outside spins faster than the inside. This makes sense, because the while wheel turns at a particular number of revolutions per minute, but if you draw a dot on the outer edge, and another dot close to the center, the dot at the outer edge will travel a greater distance than the inner dot in each rotation. It makes a bigger circle, in the same amount of time, so it must be going faster. Fast takes more force than slow, for the same weight, or, put another way, if I apply a force that accelerates the outside of a wheel to 5 mph, the inside moves much slower than 5 mph, but with the same force, which allows the inside to overcome much greater resistance.So, consider the case of a lever 1 meter long, with the pivot point 5 cm from one end. The long end must move 19 times as fast as the short end to achieve the same angular motion. If I accelerate the long end to 19 m/s with a force of 10 newtons, it will move the short end only 1 m/s, but with a force of 190 newtons. Less speed, more power.The opposite is also true. Accelerate the short end to 1 m/s, with 190 newtons of force, and the long end will be moving 19 times as fast but with 1/19th of the power.So torque is just trading speed for power.
What is the relation between engine RPM and engine torque?
Torque is the twisting or rotating force that the engine exerts on the crankshaft; it is expressed in N-m or Kg-m.It is generally measured at a radius of one meter from the crankshaft's center.Torque and Speed ( RPM) Are inversely proportional with each otherHere is the torque-speed curve.As the RPM Of an engine increases, torque increases upto a certain value of RPM. And as the RPM goes on increasing, torque starts decreasing.To understand this behaviour, you need to understand one more term which is called as “ Low-End torque”. Low -End Torque is the amount of torque produced at Low RPM range of an engine ( around 1000–2000 rpm) .This low end torque is very crucial when moving a vehicle from stand-still or driving in slow-speed conditions such as in traffic. If the greater amount of torque is generated at the lower end of engine rpm, it implies that the engine has more ‘low-end-torque’ or better pulling ability at slow speeds.It also implies that vehicle can move very quickly from rest/stand still.Now as the engine speed keeps increasing, torque starts to decrease because of the increase in friction between engine’s moving parts.Engine torque is also related to the gearing. The lower the gear, greater is the pulling ability of an engine and hence greater the torque..If you know the Horsepower (HP) of an engine, then you can calculate the torque using the formula :-Torque = (HP*5252)/RPM
How do i turn an ac or dc motor into a generator?
Im tryin to do a little home experiment. I want to know the detailed steps to turn an ac or dc motor into a power producing generator. I dont wanna pay for info so any links to free websites with details on this subject would be appreciated. Nothing complex explaining voltage and stuff like that just the actual pieces i need to add replace or remove for power production. thanx