Work Energy Method for Rotational Motion

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freshbox
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Homework Statement



Reference to part C of the question, I would like to ask for work done U1to2, how come linear motion is not included? I thought both linear and rotational exist since the parcel "move downwards" (linear) and the wheel is rotating (rotational)
Thanks...
 
Attachments
  • Printed problem text with parts (a) to (d) about a 50-kg mass F on a brake arm ABC, giving 2.1 m/s, 700 N, friction coefficient 0.3, and drum inertia 0.6 kgm^2, with mark allocations.
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  • Figure 5.1 and Figure 5.2: engineering diagrams of a rotating wheel with a hanging load F and a pivoted bar with 700 N force, labelled distances 0.65 m, 0.2 m, 0.15 m, friction coefficient 0.3, and speed v1 = 2.1 m/s.
    e2.jpg
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Last edited:
Part C asks for the work on the drum. Why worry about the falling mass? They want you to calculate the work directly.
 
Oh you you are right, sorry didn't read the question clearly :redface:

Just to confirm if they ask me to find the work of frictional force, Fr on the drum and falling mass between position 1 and 2 both linear and rotational will exist correct?

With regards to part c again, To find Torque τ, I know that τ=Tension X Radius so comparing diagram 1 (left) against diagram 2 (right) which i drew a red line to indicate Tension.

How come in the Work Done equation it uses the Frictional Force X Radius and not Tension X radius?Thanks...
 
Attachments
  • Handwritten working for a flywheel problem: a pulley diagram with a 5 kg wheel and a 20 kg hanging mass on a rope, a free-body sketch of the flywheel with tension T, torque = T x 0.015 Nm, and I = mk^2 = 5 x 0.2^2 = 0.2 kgm^2.
    c1.jpg
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  • Figure 5.1: diagram of a drum D with pivot E and radii rD = 0.25 m and rE = 0.1 m, a mass F on a cord with tension T, velocity v1 = 2.1 m/s and v2 = 0, drop distance d, a lever arm with 700 N force, friction coefficient 0.3, and 0.15 m offset labeled Frictional Force.
    c2.jpg
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freshbox said:
Just to confirm if they ask me to find the work of frictional force, Fr on the drum and falling mass between position 1 and 2 both linear and rotational will exist correct?
If the mass falls and the drum rotates, sure there will be both linear and rotational motion and kinetic energy.
With regards to part c again, To find Torque τ, I know that τ=Tension X Radius so comparing diagram 1 (left) against diagram 2 (right) which i drew a red line to indicate Tension.
Realize that those diagrams refer to different problems.
How come in the Work Done equation it uses the Frictional Force X Radius and not Tension X radius?
Part c asks you for the work done by the frictional force, not the tension.
 
My lecturer told me that τ = tension X radius this concept on the chapter of Kinetics of rotational motion diagram 1 (left)

But he never mention this on the chapter of Work Energy method for rotational motion diagram 2 (right). However, I thought they have the same meaning since both are τ in both formula.
So are you trying to say U1-2=τ X δ
The τ in the equation is actually the frictional force?

And can I say that Torque in τ=Iα and U1-2=τ X δ are actually both different thing?
 
Attachments
  • Text slide titled Kinetics of rotational motion (Newton's Second Law – Rotational Motion) giving Γ = Iα for a single torque and ΣΓ = Iα for several torques, with units Nm, kgm² and rad/s².
    ex1.jpg
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  • Slide titled Work energy method for rotational motion, with a section on work of a torque giving U1-2 = Γθ for a single torque and U1-2 = ΣΓθ for several torques, defining U, Γ and θ and listing SI units.
    ex2.jpg
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freshbox said:
My lecturer told me that τ = tension X radius this concept on the chapter of Kinetics of rotational motion diagram 1 (left)

But he never mention this on the chapter of Work Energy method for rotational motion diagram 2 (right). However, I thought they have the same meaning since both are τ in both formula.
You can calculate the torque due to any force. If the force is tangential to the wheel, then τ = Force X radius. Depending upon the particular problem, that force can be a tension force or a friction force.
So are you trying to say U1-2=τ X δ
The τ in the equation is actually the frictional force?
Right. The torque in that equation is due to the friction force. You are asked to find the work done by the friction force, so you'd use the torque created by that force to calculate the work.
And can I say that Torque in τ=Iα and U1-2=τ X δ are actually both different thing?
Yes, those are different things. The first equation is just Newton's 2nd law as applied to rotation; the torque in that formula should really be the net torque: Ʃτ=Iα.

The second equation is an expression for the work done by some torque. In your problem, it's the work done by the frictional force, which is non-conservative work that will decrease the total mechanical energy of the system.