#### Question

Obtain an expression for torque acting on a rotating body with constant angular acceleration. Hence state the dimensions and SI unit of torque.

#### Solution

a. Suppose a rigid body consists of n particles of masses m_{1}, m_{2}, m_{3}, ......, m_{n} which are situated at distances r_{1}, r_{2}, r_{3}, …, rn respectively, from the axis of rotation as shown in figure.

b. Each particle revolves with angular acceleration `alpha`.

c. Let F_{1}, F_{2}, F_{3}, …., F_{n} be the tangential force acting on particles of masses, m_{1}, m_{2}, m_{3}, …, m_{n} respectively.

d. Linear acceleration of particles of masses m1, m2,…, mn are given by, a_{1} = r_{1}α, a2 = r_{2}α, a_{3} = r_{3}α, …, a_{n} = rnα

e. Magnitude of force acting on particle of mass m1 is given by,

`F_1=m_1a_1=m_1r_1alpha` `[therefore a=r alpha]`

Magnitude of torque on particle of mass m1 is given by,

`tau_1=F_1r_1 sin theta`

But,`theta=90^@` [∵ Radius vector is ⊥ar to tangential force

`tau_1=F_1r_1sin90^@`

`=F_1r_1`

`=m_1a_1r_1`

`tau_1=m_1r_1^2alpha`

similarly

`tau_2=m_2r_2^2alpha`

`tau_3=m_3r_3^2alpha`

`tau_n=m_nr_n^2alpha`

f. Total torque acting on the body,

`tau=tau_1+tau_2+tau_3+..........+tau_n`

`tau=m_1r_1^2alpha+m_2^2alpha+m_2r_3^2alpha+......+m_nr_n^2alpha`

`therefore tau=[sum_(i=1)^nm_ir_t^2]alpha`

But `sum_(i=1)^nm_ir_i^2=I`

`therefor tau=Ialpha`

g. Unit: Nm in SI system.

h. Dimensions: `[M^1L^2T^-2]`