Question

An accelerating dv/dt=1m/s^2 is taking turn maximum possible speed vmax on a rough coefficient of friction is 0.2 horizontal circular path of radius R=10m.The value of vmax is and take g=10m/s^2


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Answer 1

Maximum possible turning speed $V_{max}=4.16(\frac{m}{s})$

Explanation:

When a particle is moving along curved path, it have tangential acceleration($a_{T}$) and radial acceleration($a_{n}$)

1. Tangential acceleration $a_{T}= 1(\frac{m}{s^{2}})$ is given

2. Radial acceleration $a_{n}= (\frac{V_{max}^{2}}{R})$

3. Total acceleration $a=\sqrt{a_{T}^{2}+a_{n}^{2}}$

4. Then inertia force (F) = mass × acceleration

                   So F= ma          ...1)

5. Now this inertia force is balance by friction force

$F=F_{f}$      ...2)

 where friction force $F_{f} = \mu \times mg$    ...3)

       

6. From equation 1, equation 2 and equation 3

$m\times \sqrt{a_{T}^{2}+a_{n}^{2}} = \mu \times mg$

$\sqrt{a_{T}^{2}+a_{n}^{2}} = \mu \times g$

$\sqrt{1^{2}+a_{n}^{2}} = \0.2 \times 10$

So $a_{n}^{2} =3$

7. On putting value radial acceleration, we got

  $\frac{V_{max}^{4}}{100}= 3$

on solving $V_{max}=4.16(\frac{m}{s})$