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3-2.Motion in Plane
medium
A uniform circular disc of radius $50\, cm$ 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 $2.0\, rad \,s^{-2}$. Its net acceleration in $ms^{-2}$ at the end of $2.0\,s$ is approximately
A$7.0$
B$6.0$
C$3.0$
D$8.0$
(NEET-2016)
Solution
Given, $r = 50 \,cm = 0.5$ m,$\alpha = 2.0\,rad\,{s^{ – 2}},{\omega _0} = 0$
At the end of $2\, s,$
Tangential acceleration, ${a_i} = r\alpha = 0.5 \times 2 = 1\,m\,{s^{ – 2}}$
Radial acceleration, ${a_r} = {\omega ^2}r = {\left( {{\omega _0} + \alpha t} \right)^2}r$
$ = {\left( {0 + 2 \times 2} \right)^2} \times 0.5 = 8\,m\,{s^{ – 2}}$
$\therefore $ Net acceleration
$a = \sqrt {a_t^2 + a_r^2} = \sqrt {{1^2} + {8^2}} = \sqrt {65} \approx 8\,m\,{s^{ – 2}}$
At the end of $2\, s,$
Tangential acceleration, ${a_i} = r\alpha = 0.5 \times 2 = 1\,m\,{s^{ – 2}}$
Radial acceleration, ${a_r} = {\omega ^2}r = {\left( {{\omega _0} + \alpha t} \right)^2}r$
$ = {\left( {0 + 2 \times 2} \right)^2} \times 0.5 = 8\,m\,{s^{ – 2}}$
$\therefore $ Net acceleration
$a = \sqrt {a_t^2 + a_r^2} = \sqrt {{1^2} + {8^2}} = \sqrt {65} \approx 8\,m\,{s^{ – 2}}$
Standard 11
Physics
