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Q.Define the angular acceleration and torque. Establish the relation between angular acceleration and torque.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 4mImportance★★★★★
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Angular acceleration α=dω/dt\alpha = d\omega/dt is how fast the angular velocity changes; torque τ=rFsin⁡θ\tau = rF\sin\theta is the moment of a force. Adding up the torques on every particle of a rigid body gives τ=Iα\tau = I\alpha, the rotational form of F=maF = ma.

1. Angular acceleration (α\alpha). When a body rotates about a fixed axis, its angular velocity ω\omega may change with time. The rate of change of angular velocity is called angular acceleration:

α=dωdt=d2θdt2\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}

Its SI unit is rad s−2^{-2} and it is a vector directed along the axis of rotation.

2. Torque (τ\tau). The torque (moment of force) about an axis is the turning effect of a force. If a force F⃗\vec{F} acts at a position r⃗\vec{r} from the axis,

τ⃗=r⃗×F⃗,τ=rFsin⁡θ\vec{\tau} = \vec{r}\times\vec{F}, \qquad \tau = rF\sin\theta

where θ\theta is the angle between r⃗\vec{r} and F⃗\vec{F}. Its SI unit is N m.

3. Relation between torque and angular acceleration. Consider a rigid body rotating about a fixed axis with angular acceleration α\alpha. Take a particle of mass mim_i at perpendicular distance rir_i from the axis. Its linear (tangential) acceleration is

ai=ri αa_i = r_i\,\alpha

so the tangential force on it is Fi=miai=miriαF_i = m_i a_i = m_i r_i \alpha, and the torque of this force about the axis is …

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