TORQUE of disks!!!!!?
Three flat disks (of the same radius) that can rotate about their centers like merry-go-rounds. Each disk consists of the same two materials, one denser than the other (density is mass per unit volume). In disks 1 and 3, the denser material forms the outer half of the disk area. In disk 2, it forms the inner half of the disk area. Forces with identical magnitudes are applied tangentially to the disks, at the outer edge of disks 1 and 2 and at the interface of the two materials for disk 3 (a) Rank the disks according to the torque about the disk center, greatest first (use only the symbols > or =, for example, 1=2>3). (b) Rank the disks according to the rotational inertia about the disk center, greatest first (use only the symbols > or =, for example, 1=2>3). (c) Rank the disks according to the angular acceleration of the disk, greatest first (use only the symbols > or =, for example, 1=2>3).
What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.3 s?
Hello Larry, you do yourself. Let me give the guidelines. OK? To determine torque you need moment of inertia. It is Mr^2/2 M = 0.24 kg. r = 0.115 m Also you need change in angular speed 9dw). That is 2 pi (d nu). d nu is 1800/60 ie 30 rev/s. You also need time dt. It is given as 4.3 s. Hence dw/dt ie angular acceleration becomes 60pi/4.3 So plug these in Torque = M r^2 /2 * 60 pi/4.3 Hope you do the plugging and simplification, The answer will be 0.07 Nm.
What torque must a motor supply to take a certain disk from 0 to 1900 rpm in 4.5 sec?
You need to be able to calculate the moment of inertia. See the reference. I = (1/2) m r² =0.5 * 0.25 kg (0.25 m)² = 0.00781 kg-m² The angular acceleration needed is 1900 rpm / 4.5 sec = 422.2 rpm/s = 422.2*2*π/60 = 44.2 rad/s² T = I α = 0.00781 * 44.2 = 0.345 N-m
Net torque; physics problem?
First, compute the angular acceleration 'a' wf = w0 + a*t where: wf, w0 = final, initial angular speeds, t=time a = wf/t = 2.38/1.46 = 1.63 rad/s**2 Next, we compute the mass moment of inertia, I = 1/2*m*r**2 for a disk I = 1/2*m*r**2 = 1/2*5.7*(.365)**2 I = .3797 kg*m**2 Finally, M = I*a = .3797*1.63 = .619 N*m
PHYSICS HOMEWORK HELP ANSWER AS MANY AS U CAN THANkS : )?
i need to show work also guy thanks a lot for people answering these questions 46.a fixed 0.15 kilogram solid-disk pulley with a radius of 0.075 is acted on by a net torque of 6.4m.N. what is the angular acceleration of the pulley? 47. what net torque is needed to give a uniform 20-kilogram solid ball with a radius of .20m and angular acceleration of 2.0rad/s^2 52. A 10 kilogram solid disk of radius .50m is rotated about an axis through its center.If the disk accelerates from rest to an angular speed of 3.0rad/s while rotating 2.0 recolutions,what net torque is required?
Physics help?
The 20-cm-diameter disk in the figure can rotate on an axle through its center. What is the net torque about the axle? Figure can be seen at: http://www.brainmass.com/homework-help/physics/modern-physics/54849 and click view attach file. I know you have to add all the torques together, but I can't get the answer. Please provide answer, and how to do it. Thank You.
Physics Homework Help! :)?
Use Torque(total sum) = I*a where I = inertia and a = angular acceleration or dw/dt where w is angular velocity Use I = 1/2m*v^2 They give you alpha, or a, so find total torque T1 = 1800N*.35m T2 = -22Nm T1 - T2 = 608 607N*m = I(90rad/s^2) You have the inertia And inertia = 1/2mv^2, don't confuse v with w, you must account for the radius to transform into circular motion.
If no external torque acts on the body, will the angular velocity be conserved?
The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occurof course not ... torque is the rate of change of angular momentum and angular momentum in simple sense can be seen as the product of angular velocity with moment of inertia. Changing angular velocities are behind spin skating, toe dancing etc.example: on a disk M and radius R rotating with angular speed ωi drops a small object of mass m on the edge of disk before Mif=Iiωi=1/2MR^2ωi after Mf=Ifωf=(1/MR^2+mR^2)ωf it must be 1/2MR^2ωi=(1/MR^2+mR^2)ωf then ωf=ωi(1/2MR^2)/(1/MR^2+mR^2).♣♥♠
How do I calculate effective rotor radius for a disc in brakes for a car?
Thanka for A2A. I think this will help:For more queries,Brake System Design Calculations
Why do tractors have big rear wheels and small front wheels?
Hi..Greetings to all.. While I respect, all of you, Yes, the answers from all of you are right, but partly.You need a bigger tire, just because, the tractor is operated in wet sludgy crop fields, where your normal size tire cannot work since, normal tire will simply keep spinning without any plough action.In technical terms, the need for bigger rear wheel in tractor is not just one , but many.. they are1. The Broader, bigger wheel with a Thick Lug Pattern in the Tires will help for a very good contact pattern with the surface, in this case, mostly the crop fields are expected to be wet and slippery. Hence, Bigger broader wheel for better contact with the ground surface, which will inturn enable the tractor to plough the field better. In contrast, the Front wheels are made small , comparatively slim and only intermittent rib pattern in the tires, this will help for ease of steering even in wet sludgy surfaces.2. Most importantly, the bigger wheel enables, addition of Dead Weights on its Rim Plate, (hope u have seen the rear wheels with dead weights), these deadweights are added for higher grip with the wet sludgy field where the tractor is operated.3. Higher Ground Clearance is achieved with a bigger wheel, so as to pass over the plants in the field.4. Regarding Torque / other info... It's very simple, dont get too complicated. The bigger the Tire, more is the torque need to rotate the tires, which means, you will need bigger engines, inturn, there will be higher fuel consumption and the cost of tractor becomes heavy. So, from a design perspective, no tractor is designed to intentionally make them big, unless there is a need for it. As such, the Torque of the engine being a reason for the big size tire is ruled out.Thanks,