Calculating Torque for a Robot Wheel: Mass, Acceleration, and Constant Velocity

  • Thread starter Thread starter elimenohpee
  • Start date Start date
  • Tags Tags
    Torque Wheel
AI Thread Summary
To calculate the torque needed for a robot wheel with a mass of 3 kg and a diameter of 0.0381 m, the correct formula involves using torque = r × F, where F is the force applied. The initial calculation of torque as 1.12 Nm is questioned, as it may seem high for the mass involved. When considering acceleration at 1.47 m/s², the torque must account for this additional force. For constant velocity at 5 m/s, the torque calculation simplifies, as acceleration is zero, focusing solely on overcoming friction and inertia. Understanding the correct application of torque principles is crucial for accurate calculations in robotics.
elimenohpee
Messages
64
Reaction score
0

Homework Statement


You are given a robot, and you are asked to design the torque needed for the driving motor. What is the torque to turn a wheel who's mass of the system connected to the rim is 3kg, and the diameter of the rim is 0.0381m?

What is the torque when the wheel accelerates at a rate of 1.47m/s^2? What is the torque when the wheel reaches a constant velocity of 5 m/s?


The Attempt at a Solution



I know torque = force * distance = mass * acceleration * distance

I figured the torque to turn the wheel would be:
torque = m*g*r = 3*9.81*0.0381 = 1.12Nm (but this sounds like a lot for such a small mass)

but I'm not sure what to do when the wheel is to accelerate. And I'm lost what to do when the acceleration is zero, aka constant velocity.
 
Physics news on Phys.org
elimenohpee said:
I know torque = force * distance = mass * acceleration * distance

I figured the torque to turn the wheel would be:
torque = m*g*r = 3*9.81*0.0381 = 1.12Nm (but this sounds like a lot for such a small mass)

but I'm not sure what to do when the wheel is to accelerate. And I'm lost what to do when the acceleration is zero, aka constant velocity.

sorry but...your torque equation is wrong
remember that torque is r X F not r * F
r X F is also rFsin(theta), with theta as the angle between them
also that r is not the diameter of the wheel
it's the distance from the point of force applied
since it is a wheel, u can tpretty much use the center for axis of rotation and the distance to where it spins is yyour r
 
Yes I'm aware that, the angle between the vectors was assumed to be pi/2. And that distance given was the radius of the wheel, since the force is assumed to applied at the center of the rim.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top