(P.15) Angular Measure expressed in radians

AI Thread Summary
Equation D, ω = Vt/r, is unique because it relates angular velocity to linear distance, requiring the angle to be measured in radians for the equation to hold true. The radian is defined such that the arc length equals the radius multiplied by the angle in radians, making it essential for this relationship. Other equations discussed, such as α = Δθ / Δt and ω = Δω / Δt, do not require radians since they involve angular measures that can be in any consistent unit. The distinction lies in the fact that Equation D combines angular and linear quantities, necessitating the use of radians for accurate calculations. Understanding this relationship clarifies why radians are crucial in certain rotational dynamics equations.
gcombina
Messages
157
Reaction score
3
Which equation is valid only when the angular measure is expressed in radians?
a) α = Δθ / Δt

b) ω= Δω / Δt

c) ω^2 = ωo^2 + 2αθ

d) ω = Vt/r (here T is a subscript)

e) θ = 1/2αt^2 + ωαt

Answer is D but why??

* I am totally lost so I can not show work
 
Last edited:
Physics news on Phys.org
So equation D is saying that angular velocity multiplied by the radius is equal to tangent velocity.

Think of an object on a string spinning around in a circle. The length of the string is r (which is the radius of the path of the object). The angular velocity is \frac{Δθ}{Δt} but the tangent velocity is \frac{length.of.arc .subtended.by.Δθ}{Δt}

Multiplying an angle by the radius will give you the length of the arc subtended by that angle ONLY if the angle is measured in radians (that's the definition of the radian, and the reason it's such a useful unit)
That is why that statement is only true if it is measured in radiansHope this was clear
 
gcombina said:
a) α = Δθ / Δt

b) ω= Δω / Δt

Also shouldn't it be:?

a) ω = Δθ / Δt

b) α = Δω / Δt
P.S.
The reason the other equations don't need to be measured in radians is that they just involve angles, so it doesn't matter if it's radians, degrees, or something else, because the answer will come out in the same angular units.

Equation D is the only one that deals with angles AND (linear) distances.
 
"The reason the other equations don't need to be measured in radians is that they just involve angles, so it doesn't matter if it's radians, degrees, or something else, because the answer will come out in the same angular units."
(so if something is measured in degrees, the answer will be defaulted into radians?)

Equation D is the only one that deals with angles AND (linear) distances". (I think when something is measured in angles then we have degrees right? so if you say that equation D only deals with angles and linear distances, isn't angles always measured in degrees?)



PS:
Equation D shows angular velocity which is by default measured in radians right? basically that is the hint that I got but still a little confused with these equations.
 
gcombina said:
Nathanael said:
"The reason the other equations don't need to be measured in radians is that they just involve angles, so it doesn't matter if it's radians, degrees, or something else, because the answer will come out in the same angular units."
(so if something is measured in degrees, the answer will be defaulted into radians?)
No. Nathanael is saying that in an equation like ω = dθ/dt it doesn't matter whether the angles are measured in radians or degrees as long as you are consistent. It's exactly the same with linear distances; the equation v = ds/dt is valid in various units as long as they match: m/s, m and s, or mph, miles and hours, etc. If you were to change units it would multiply both side of the equation by the same factor.
Nathanael said:
Equation D is the only one that deals with angles AND (linear) distances".
(I think when something is measured in angles then we have degrees right?)
No, you can measure angles in degrees or radians, or revolutions, or whatever you care to define.
I take a slightly different view from Nathanael's. To me, what's special about equation D is that on the RHS you have a velocity divided by a distance which is not in the same direction. Indeed, it's at right angles.
Suppose I'm going in a circle radius r at a steady speed v. After time t, I've gone a distance vt. If I divide that by the radius of the circle I get a measure of how far round I've gone, so vt/r is giving that angle in some units. When I've gone distance 2πr I will have gone once around the circle, so the units must be such that an angle of 2π in those units is 360 degrees. This serves to define the unit radian.
 
Also note that it is important to learn to use subscripts. Vt is the tangential velocity while Vt is just the velocity times time. These are two different things. Use the go advanced button and look for the subscript button that looks like a X2.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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 hanging 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