Question

prove that: z =r F s in Q. with the help of this expression find condition for a torgue to be maximum and minimum​

Answer 1

Answer:

= F × r × sinθ

T = torque

F = linear force

r = distance measured from the axis of rotation to where the application of linear force takes place

theta = the angle between F and r

In this formula, sin(theta) has no units, r has units of meters (m), and F happens to have units of Newtons (N). Combining these together, one can see that a unit of this force is a Newton-meter (Nm).

The Formula Derivation

The SI unit for torque happens to be the newton-meter (N⋅m).

Now let’s find the formula or expression.

Rate of change of Angular Momentum in relation to time = ΔL/ΔT

Now, ΔL/ΔT = Δ(I ω)/ΔT = I. Δω/ΔT ……. (1) Here I is certainly the constant when mass and shape of the object are unchanged

Now Δω/ΔT refers to the rate of change of angular velocity with time i.e. angular acceleration (α).

So from equation 4 one can write, ΔL/ΔT = I α …………………(2)

I (moment of inertia) refers to the rotational equivalent of mass(inertia) of linear motion. Similarly, angular acceleration α (alpha) certainly refers to the rotational motion equivalent of linear acceleration.

So from equation 5 one can get, ΔL/ΔT = τ ……………………. (6) this certainly states that the rate of change of angular momentum with time is called Torque.

Torque (T) refers to the moment of force. Τ = r x F = r F sinθ ……………. (3)

F is the force Vector and r refers to the position vector

θ happens to be the angle between the force vector and the lever arm vector. ‘x’ certainly denotes the cross product.

Τ = r F sin θ = r ma sinθ = r m αr sinθ =

m

r

2

. α sinθ = I α sinθ = I X α ……………………… (4)

[α is angular acceleration, I refers to the moment of inertia and X denotes cross product.]

T = I α (from equation 4)

or, T = I (ω2-ω1)/t [here α = angular acceleration = time rate of change of the important angular velocity = (ω2 – ω1)/t where ω2 and ω1 happen to be the final and initial angular velocities and t is the time gap]

or, T t = I (ω2-ω1) ……………………(5)

when, T = 0 (i.e., net torque is zero),

I (ω2-ω1) = 0

i.e., I ω2=I ω1 ………….. (6)

Solved Example on Torque Formula

Q1. A car mechanic applies a force of 800 N to a wrench for the purpose of loosening a bolt. He applies the force which is perpendicular to the arm of the wrench. The distance from the bolt to the mechanic’s hand is 0.40 m. Find out the magnitude of the torque applied?

Answer: The angle between the moment the arm of the wrench and the force is without a doubt 90°, and sin 90° θ = 1. The torque is:

T = F × r × sinθ

Therefore, magnitude of the torque = (800N) (0.4m) = 320 N∙m

Hence, the magnitude of the torque is 320 N