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Damm [24]
3 years ago
7

The angular velocity of the disk is defined by ω = ( 5 t 2 + 2 ) r a d / s , where t is in seconds. Determine the magnitudes of

the velocity and acceleration of point A on the disk when t = 0.5 s .
Physics
1 answer:
ohaa [14]3 years ago
6 0

Answer

given,

Assume radius of the disk be = 0.8 m

angular velocity of disk

ω = ( 5 t² + 2 ) r a d / s

magnitude of velocity and acceleration = ?

At the instant of time, t = 0.5 s

ω = ( 5 x (0.5)² + 2 ) r a d / s

ω = 3.25 r a d / s

\alpha = \dfrac{d\omega}{dt}

\alpha = \dfrac{d}{dt}( 5 t² + 2)

\alpha =10 t

\alpha =10\times 0.5

α = 5 rad/s²

velocity

v = ω r

v = 3.25 x 0.8

v = 2.6 m/s

tangential acceleration

a_t = \alpha r

a_t =5 \times 0.8

a_t =4\ m/s^2

normal acceleration

a_n = \omega^2\ r

a_n = 3.25^2\times 0.8

a_n =8.45 \m/s^2

a = \sqrt{a_n^2 + a_t^2}

magnitude of the acceleration

a = \sqrt{8.45^2 + 4^2}

a = 9.35 m/s²

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final speed of the elevator, v = 2.05 m/s

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a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2

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