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Question-225394




Question Number 225394 by fantastic last updated on 24/Oct/25
Commented by fantastic last updated on 24/Oct/25
you are right sir.  please show your work sir.
$${you}\:{are}\:{right}\:{sir}. \\ $$$${please}\:{show}\:{your}\:{work}\:{sir}. \\ $$
Commented by mr W last updated on 24/Oct/25
i got t=((3ωR)/(4gk))
$${i}\:{got}\:{t}=\frac{\mathrm{3}\omega{R}}{\mathrm{4}{gk}} \\ $$
Answered by mr W last updated on 24/Oct/25
say m=mass of disc  ρ=(m/(πR^2 ))  τ=∫_0 ^R 2πr×kρg×rdr    =((2kmg)/R^2 )×∫_0 ^R r^2 dr=((2kmgR)/3)  I=((mR^2 )/2)  Iα=τ  ((mR^2 α)/2)=((2kmgR)/3)  ⇒α=((4gk)/(3R))  ⇒t=(ω/α)=((3ωR)/(4gk)) ✓
$${say}\:{m}={mass}\:{of}\:{disc} \\ $$$$\rho=\frac{{m}}{\pi{R}^{\mathrm{2}} } \\ $$$$\tau=\int_{\mathrm{0}} ^{{R}} \mathrm{2}\pi{r}×{k}\rho{g}×{rdr} \\ $$$$\:\:=\frac{\mathrm{2}{kmg}}{{R}^{\mathrm{2}} }×\int_{\mathrm{0}} ^{{R}} {r}^{\mathrm{2}} {dr}=\frac{\mathrm{2}{kmgR}}{\mathrm{3}} \\ $$$${I}=\frac{{mR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}\alpha=\tau \\ $$$$\frac{{mR}^{\mathrm{2}} \alpha}{\mathrm{2}}=\frac{\mathrm{2}{kmgR}}{\mathrm{3}} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{4}{gk}}{\mathrm{3}{R}} \\ $$$$\Rightarrow{t}=\frac{\omega}{\alpha}=\frac{\mathrm{3}\omega{R}}{\mathrm{4}{gk}}\:\checkmark \\ $$
Commented by fantastic last updated on 24/Oct/25
What dote do you see  next to my comment  sir.  24 /25 pls tell
$${What}\:{dote}\:{do}\:{you}\:{see} \\ $$$${next}\:{to}\:{my}\:{comment} \\ $$$${sir}. \\ $$$$\mathrm{24}\:/\mathrm{25}\:{pls}\:{tell} \\ $$

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