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 m1, m2, m3, ......, mn which are situated at distances r1, r2, r3, …, rn respectively, from the axis of rotation as shown in the figure.
b. Each particle revolves with angular acceleration `alpha`.
c. Let F1, F2, F3, …., Fn be the tangential force acting on particles of masses, m1, m2, m3, …, mn respectively.
d. Linear acceleration of particles of masses m1, m2,…, mn are given by, a1 = r1α, a2 = r2α, a3 = r3α, …, an = 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]`
