Question

Q] Obtain the expression for the torque acting on a relating body with constant angular acceleration.​

Answer 1

Answer:

$\huge \mathbb\fcolorbox{cyan}{cyan}{Answer}$

A system rotating with an angular momentum in presence of a torque suffers a change in the angular momentum and the rate of change of angular momentum is directly proportional to the torque acting on it.

If I be the moment of inertia of the system and ω be the angular velocity then angular momentum L= Iω

Rate of change of angular momentum provided the shape of system does not change is :

τ= dt÷d = (Iω)=I×dω÷dt = Iα

α is the angular acceleration and τ is the torque.Dimension of Torque is [ML²T-2²] and its unit is Newton- metre (N⋅m).

Explanation:

I hope this will be correct and it will help you ✨⚡

Answer 2

Q] Obtain the expression for the torque acting on a rotating(not relating)body with constant angular acceleration.

Solution:

Let's consider a rotating body having n number of particles at different radius.

Let their masses be $m_1, m_2, m_3, m_4,------,m_n$ and radius at which they are located be $r_1, r_2, r_3, r_4-------r_n$

Now, we know that Torque( $\tau$ ) = Force x radius

Calculating the torque of individual particle of the rotating body,we get:

$\tau =\tau_1 + \tau_2+\tau_2+\tau_3+\tau_4+. +. +. +. +. +\tau_n \\\\ \tau=F_1r_1+F_2r_2+F_3r_3+F_4r_4+. +. +. +. +. +F_nr_n$

Let's consider that the particles of the rotating body has their individual accelerations $a_1, a_2, a_3, a_4,-------,a_n$

Replace F by ma

$\tau=m_1a_1r_1+m_2a_2r_2+ m_3a_3r_3+ m_4a_4r_4+. +. +. +. +. +m_na_nr_n$

Now, since the object is rotating, we will take into consideration, the tangential acceleration which is represented by $r \alpha$ where r is the radius and $\alpha$ is the angular acceleration which is constant for all the particles of the body.

a= $r \alpha$

Now,

$\tau=m_1r_1r_1 \alpha, m_2r_2r_2 \alpha+m_3r_3r_3 \alpha+m_4r_4r_4 \alpha++. +. +. +. +. +,m_nr_nr_n \alpha$

Taking $\alpha$ common, we get:

$\tau= \alpha(m_1r_1^2+m_2r_2^2+m_3r_3^2+m_4r_4^2+. +. +. +. +. +m_nr_n^2)$

$\tau=\alpha \bigg( \sum_{t=1}^{t=n} ~m_tr_t^2 \bigg)$

But, $\sum_{t=1}^{t=n} ~m_tr_t^2$ is moment of Inertia denoted by I

So, the final relation we get is $\huge{\bf{\tau= \alpha \times I}}$