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A uniform circular disc of radius $\mathbf{5 0} \mathbf{~ c m}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0 \mathrm{rad} \mathrm{s}^{-2}$. Its net acceleration in $\mathrm{ms}^{-2}$ at the end of $2.0 \mathbf{s}$ is approximately: A $\quad 8.0$ B $\quad 7.0$ C $\quad 6.0$ D $\quad 3.0$
A uniform circular disc of radius $\mathbf{5 0} \mathbf{~ c m}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0 \mathrm{rad} \mathrm{s}^{-2}$. Its net acceleration in $\mathrm{ms}^{-2}$ at the end of $2.0 \mathbf{s}$ is approximately:
A $\quad 8.0$
B $\quad 7.0$
C $\quad 6.0$
D $\quad 3.0$
Physics
A uniform circular disc of radius $\mathbf{5 0} \mathbf{~ c m}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0 \mathrm{rad} \mathrm{s}^{-2}$. Its net acceleration in $\mathrm{ms}^{-2}$ at the end of $2.0 \mathbf{s}$ is approximately:
A $\quad 8.0$
B $\quad 7.0$
C $\quad 6.0$
D $\quad 3.0$

Correct Option A

Solution:

As we know, Torque $\tau=\mathrm{I} \alpha$

Given, $\alpha=2 \mathrm{rad} / \mathrm{s}^{2}$

As we know, Tangential acceleration $\mathrm{a}=\mathrm{r} \alpha$

$\mathrm{a}_{t}=\frac{1}{2} \times 2=1 \mathrm{~ms}^{-2}$

$\mathrm{a}_{t}=1 \mathrm{~ms}^{-2}$

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$=0+2$

$\mathrm{v}=2 \mathrm{~m} / \mathrm{s}$

Also, Radial acceleration

$\mathrm{a}_{\mathrm{r}}=\frac{\mathrm{v}^{2}}{\mathrm{r}}=\frac{4}{0.5}=8 \mathrm{~ms}^{-2}$

Thus, calculating net acceleration

$\mathrm{a}=\sqrt{\mathrm{a}_{\mathrm{r}}^{2}+\mathrm{a}_{\mathrm{t}}^{2}}$

$=\sqrt{64+1}$

$=8 \mathrm{~ms}^{-2}$ (approx)

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