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  • Two hoops, each of mass  and radius ' <img alt="2 \mathrm{~m}" src="https://entra

Two hoops, each of mass 1 \mathrm{~kg} and radius ' 2 \mathrm{~m} ' are joined with a hollow cylindrical tube of negligible mass, and radius ' 1 \mathrm{~m} ' with the help of spokes as shown in the figure.

Assuming friction is sufficient to avoid slipping, find the acceleration of the block of mass ' \mathrm{1kg} ' which is connected with the cylindrical tube by a string of negligible mass, and going over an ideal pulley.

Option: 1

2 \mathrm{~m} / \mathrm{s}^2


Option: 2

\frac{20}{3} \mathrm{~m} / \mathrm{s}^2


Option: 3

\frac{10}{3} \mathrm{~m} / \mathrm{s}^2


Option: 4

\frac{10}{9} \mathrm{~m} / \mathrm{s}^2


Answers (1)

best_answer

As the hoop is not slipping on the surface-

for point A:-

    \begin{aligned} v_2 & =r_2 \omega \\ \omega & =\frac{v_2}{r_2} \end{aligned}

As the string is inextensible -

\begin{aligned} & v_1=v_2-r_1 \omega=v_2-r_1\left(\frac{v_2}{r_2}\right) \\ & v_1=v_2\left[1-\frac{r_1}{r_2}\right] \\ & v_1=v_2\left[1-\frac{1}{2}\right]=\frac{v_2}{2} \mathrm{~m} / \mathrm{s} \end{aligned}

Applying conservation of energy when the block descend by ' x ' and having velocity v_1
    \begin{aligned} g x & =\frac{1}{2} \times 1 \times v_1^2+1 \times v_2^2 \\ g x & =\frac{1}{2} v_1^2+4 v_1^2=\frac{g}{2} v_1^2 \\ v_1 & =\sqrt{\frac{2 g}{9} x} \end{aligned}

Applying third equation of motion:-

    \begin{aligned} & v^2=u^2+2 a s \\ & v_1^2=0+2 a \times x \\ & \frac{2 g}{9} x=2 a x \\ & a=\frac{g}{9}=\frac{10}{9} \mathrm{~m} / \mathrm{s}^2 \end{aligned}

Posted by

seema garhwal

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