# Obtain an Expression for Torque Acting on a Rotating Body with Constant Angular Acceleration. Hence State the Dimensions and Si Unit of Torque. - Physics

Sum

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]

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 1 Rotational Dynamics
Exercises | Q 8 | Page 24