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  • A disc of mass '' rests on a plank of mass '<img alt="1 \mathrm{~kg}" s

A disc of mass '10 \mathrm{~kg}' rests on a plank of mass '1 \mathrm{~kg}' as shown in the figure.

The coefficient of friction is '0.1' for all the surfaces in contact. Initially the system was in rest, a horizontal force of 100 \mathrm{~N} is applied on the plank as shown in the figure. What can be said regarding the motion of the disc over the plank. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

Disc will roll on the plank


Option: 2

Disc will initially slide then starts rolling afterwards


Option: 3

Under given situations, disc will never roll on the plank


Option: 4

Both will remain at rest


Answers (1)

best_answer

For Disc

N_1=M g\\ f_{\text {max }}=\mu N_1=0.1 \times 10 \times 10=10 \mathrm{~N}


f^{\prime}=\mu(m+M) g=0.1 \times 11 \times 10=11 \mathrm{~N}
For translatory motion of the disc
f=M a_{CM}~~~~~~~~~~~~...(i)

For rotational motion of the disc
\begin{aligned} & f r=\frac{M r^2 \alpha}{2} \\ & f=\frac{M}{2}(r \alpha) \quad ~~~~~~~~~~~~\ldots \text { (ii) } \end{aligned}

From equation (i) and (ii)
a_{CM}=\frac{r \alpha}{2}
For the translator motion of the plate-
\begin{aligned} F-\left(f+f^{\prime}\right)&=m a_p \\ 100-21&=1 \times a_p \\ a_P&=79 \mathrm{~m} / \mathrm{s}^2 \end{gathered}
Acceleration of point of contact of the disc placed on the plank.
\begin{aligned} & a=a_{CM}+r \alpha=\frac{3 r \alpha}{2} \\ & a=3 a_{CM} \\ & a=3\left(\frac{f}{M}\right)=3\left(\frac{10}{10}\right)=3 \mathrm{~m} / \mathrm{s}^2 \end{aligned}

As, \quad a_p>a, sliding will take place.

Posted by

Devendra Khairwa

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