Torque And Angular Acceleration Help Eee

Angular acceleration, RPM and torque help needed?

An aircraft sits on a runway ready for take off. It has 1.4m diameter wheels and accelerates uniformly from rest to 225km/hr (which is the take off velocity) in 40s. Determine;
-angular acceleration of the undercarriage wheels?
-the number of revolutions made by each wheel during the take off run?
-the torque exerted on each wheel if the mass is 85kg and the effective radius of gyration is 45cm?
Any help with workings is greatly appreciated
Cheers

ANGULAR ACCELERATION AND TORQUE?

I think the experiment you performed was to measure the angular acceleration of a flywheel produced by a torque provided by a mass on a string wound round the wheel. Torque is the independent variable (the thing you change, by changing the mass), angular acceleration is the dependent variable (the thing you are trying to measure).

The torque T of a pair of forces is one force x perpendicular distance between them. Here T=WR where W is the weight of the falling mass and R is the radius of the wheel. (The second force in the pair is the reaction at the axle of the wheel.)

W=mg is a force. Torque is force x perpendicular distance = WR = mgR, not mgR^2.

The "a" in the equation a = 2S/t^2 is the LINEAR acceleration in m/s^2 of the falling mass m. When the mass has fallen a distance S then the cylinder has rotated S/2πR times and has turned through an angle θ = S/R radians (since 1 rev = 2π rad). The ANGULAR acceleration α = d2θ/dt2 = a/R.

Torque and angular acceleration are related by

T = moment of inertia (kg m^2) x angular acceleration α (rad/s^2).

So you will plot either α = a/R = 2S/Rt^2 (y axis) against T=mgR (x axis), or 1/t^2 against m, and expect to get a straight line whose slope (in the first case) equals the moment of inertia of the flywheel. You can check if slope does = MI of wheel; if the wheel is a solid cylinder of radius R, the MI is ½MR^2 where M is its mass.

Torque and angular acceleration problem....please help me?

the force will generate a torque given by

torque = Fr sin theta where F is the magnitude of the force (100N), r is the distance between the application of the force and pivot (0.3m), and theta is the angle between the force and radius vector (90 degrees since the force is tangent to a circular disk)

therefore the torque in this case is 100N x 0.3m = 30 Nm

this torque will generate an angular acceleration:

Torque = I A where I is the moment of inertia and A is the angular acceleration (as opposed to "a" the linear acceleration)

I for a disk is 1/2 MR^2, so for this disk we have

I= 1/2( 4kg)(0.3m)^2 = 0.18kgm^2

so the angular acceleration is

A=torque/I = 30Nm / 0.18 kgm^2 = 166.67 rad/s/s

linear acceleration, a, is related to angular acceleration through:

a = r A; substitute values and solve for a

b) the torque is the same since you have the same pulling force and radius; the difference is that the moment of inertia is different

the moment of inertia of a hoop or cylinder is MR^2, so here the value of I is twice its value in part a); therefore the angular acceleration is

A=Torque / I and will be half the previous value

similarly, the linear acceleration will be halved

Physics and Torque and Angular Acceleration, Please HELP?

How are they applied? Tangentially? In opposite direction? Or same?
Let us assume oppo direc and on the border.
Then net force=35N.
Torque = r x F = I w
=35x0.341= ''0.5'' m R^2
R radius m mass
''Check if it is 0.5 only for a solid disk or cyliner about its axis.

Hope it helps

Torque / angular acceleration question (physics)?

Solution:
1) Identify:
We assume the rod is circular with even diameter. And it will not be influenced by any other forces than weight and the contact force in A. In addition there is frictionless pivot in A.

2) Set up:
The way to handle this problem is to first find the torque (around A) due to weight of rod. Second is to use the relationship between torque and angular acceleration and third is to find the tangential acceleration for the rod in B using the relation between angular acceleration and tangential acceleration.

3) Execute:
3.1) Torque (Tau) = Weight (G) x 0.5 x Length (L) = 2.1 [kg] x 9.81 [m/s^2] x 0.5 x 0.61 [m] = 6.28 [Nm].
Tau=Moment of Inertia (I) x angular acceleration (alpha).

3.2) Moment of Inertia (I):
I = (1/3) x mass of rod (M) x L^2 = (1/3) x 2.1 [kg] x (0.61 [m])^2 = 0.26 [kgm^2].

3.3) Angular acceleration (alpha):
alpha = tau/I =(6.28 [Nm]/0.26 [kgm^2]) = 24.15 [rad/s^2].

3.4) Tangential acceleration (a_t):
Relationship between tangential acceleration (a_t) and angular acceleration (alpha) is; a_t = L x alpha = 0.61 [m] x 24.15 [rad/s^2] = 14.7 [m/s^2].

4) Evaluate:
As seen the initial tangential acceleration, a_t > g (gravitational acceleration). In fact the relationship is; a_t = (3/2) x g. Is this correct?
You can check by doing a simple experiment.Make arrangement as described in (Identify) and place a coin in A. Release rod and observe behavior of coin.

Physics Problems Torque, Angular Acceleration?

If the shaft is horizontal, the frictional force varies with angle. It is max at θ = 90º, is zero at 0º and 180º and is negative from 180º to 360º. The average normal force in the upper circle half is then

(1/π)*∫m*g*sinθ*dθ [from θ = 0 to θ = π) = -(1/π)*m*g*cosθ [θ = 0 to θ = π] = -(1/π)*m*g*(-1 - 1) = (2/π)*m*g

The friction force is µ*m*g*(2/π) (m is the flywheel mass)

The friction torque is µ*m*g*(2/π)*Rs (Rs is the shaft radius)

The angular acceleration is T/I, where I = moment of inertial of the flywheel

I = 0.5*m*(Rf² - Rs²) (Rf is the flywheel radius)

Angular acceleration is then α = µ*g*(4/π)*Rs/(Rf² - Rs²)

The angular velocity is ω = α*t

t = ω/α The time to go from ω = 0 to ω = 360 RPM = 37.7 rad/s is then 37.7/α s

Torque and angular acceleration problem...Keep getting the wrong answer...Please Help!?

we write the equations of motion for the stone and pulley

the forces acting on the stone are the tension in the wire and the weight of the stone, so we have

T - mg = -ma m=mass stone and a=linear accel

this tension causes a torque of magnitude TR2 where R2 is the outer radius of the pulley (and R1 is the inner radius)

the torque causes an angular acceleration (A) given by

torque = I A where I is the moment of inertia of the pulley and A is the angular acceleration

for this pulley, the moment of inertia is
I=1/2 M(R1^2+R2^2) = 1/2 (10kg)(0.6^1+1^2)
I=6.8kgm^2

the angular acceleration is related to the linear acceleration according to

a= R2 A

putting these together, we have

TR2 = 6.8 (a/R2) or T=6.8a/R2^2 = 6.8a

substitute this into the first equation:

T-mg=-ma

T=mg-ma
6.8a = 2g-2a
a=2g/8.8 = 2.2m/s/s

T = m(g-a)=2kg(9.8-2.2)m/s/s = 15.1N

A=a/R2=2.2rad/s/s

What is the relationship between torque and acceleration?

This is really a very simple question. First, we must note that angular accel. X radius=accel. Torque is Force times length of the axis. In linear mechanics, we have dP/dt=F net, in other words F=m times accel. This would simplify to Torque/axis length=m*radius times angular accel. This simplifies to Torque=Mass*radius*angular accel*axis length or Torque=Mass*r^2*alpha. Thus as torque increases angular accel. (alpha ) shall also increase I think BOGATU"S EXPLANATION IS MORE OR LESS CORRECT< HOWEVER HE DOES NOT GIVE THE ANSWER