A-circular-disc-of-mass-20kg-and-radius-15cm-is-mounted-in-an-horizontal-cylinder-axel-of-radius-1-5cm-Calculate-the-kinectic-energy-of-the-disc-after-1-2-secs-if-a-force-of-12N-is-applied-tangential

Question Number 10640 by Saham last updated on 21/Feb/17

A circular disc of mass 20 kg and radius 15 cm is mounted in an

horizontal cylinder axel of radius 1 . 5 cm, Calculate the kinectic

energy of the disc after 1.2 secs if a force of 12 N is applied tangentially

to the axel .

Answered by mrW1 last updated on 21/Feb/17

m o m e n t o f i n e r t i a I = \frac{1}{2} {m r}^{2} = \frac{1}{2} \times 20 \times 0 . 15^{2} = 0.225 {k g m}^{2}

t o r q u e T = {F r}_{1} = 12 \times 0.015 = 0.18 N m

w i t h α = a n g u l a r a c c e l e r a t i o n

T = I α

\Rightarrow α = \frac{T}{I} = \frac{0.18}{0.225} = 0.8 r a d / s^{2}

a n g u l a r v e l o c i t y

ω = α t = 0.8 \times 1.2 = 0.96 r a d / s

k i n e t i c e n e r g y

K E = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 0.225 \times 0 . 96^{2} = 0.10368 J

o r

K E = w o r k = F s = F θ r_{1}

θ = \frac{1}{2} α t^{2} = \frac{1}{2} \times 0.8 \times 1 . 2^{2} = 0.576 r a d

\Rightarrow K E = 12 \times 0.576 \times 0.015 = 0.10368 J

Commented by Saham last updated on 21/Feb/17

wow . i really appreciate sir . God bless you .

Answered by remember last updated on 21/Feb/17

K = W = F s

s = \frac{1}{2} α {r t}^{2} = \frac{1}{2} (\frac{F r}{{M R}^{2} / 2}) t^{2} r

S o K = F s = {(\frac{F t r}{R})}^{2} \frac{1}{M}

= {(\frac{12 \times 1.2 \times 0.015}{0.15})}^{2} (\frac{1}{20})

= 0.10368 J

Commented by remember last updated on 21/Feb/17

Commented by Saham last updated on 21/Feb/17

Great, i really appreciate .

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