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 the 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"_1"a"_1 = "m"_1"r"_1alpha` `[therefore "a"="r" alpha]`

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

`tau_1="F"_1"r"_1 sin theta`

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

`tau_1="F"_1"r"_1sin90^@`

`="F"_1"r"_1`

`="m"_1"a"_1"r"_1`

`tau_1="m"_1"r"_1^2alpha`

similarly

`tau_2="m"_2"r"_2^2alpha`

`tau_3="m"_3"r"_3^2alpha`

`tau_"n"="m"_"n""r"_"n"^2alpha`

Total torque acting on the body,

f. `tau=tau_1+tau_2+tau_3+..........+tau_"n"`

`tau="m"_1"r"_1^2alpha+"m"_2^2alpha+"m"_2"r

"_3^2alpha+......+"m"_"n""r"_"n"^2alpha`

`therefore tau=[sum_(i=1)^"n""m"_"i""r"_"t"^2]alpha`

But `sum_("i"=1)^"n""m"_"i""r"_"i"^2="I"`

`therefore tau=Ialpha`

g. Unit: Nm in SI system.

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