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Obtain an Expression for Torque Acting on a Rotating Body with Constant Angular Acceleration. Hence State the Dimensions and Si Unit of Torque. - HSC Science (General) 11th - Physics

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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 m1, m2, m3, ......, mn which are situated at distances r1, r2, r3, …, rn respectively, from the axis of rotation as shown in 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_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]`

 

 

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Solution Obtain an Expression for Torque Acting on a Rotating Body with Constant Angular Acceleration. Hence State the Dimensions and Si Unit of Torque. Concept: Torque and Angular Momentum.
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