Question

A uniform slender bar $AB$ of mass $m$ is suspended, as shown in figure, from a disc of the same mass $m$. Determine the accelerations of points $A$ and $B$ immediately after a horizontal force $F$ has been applied at $B$. Assume that the disc rolls without sliding on the wedge.

• A. $\frac{2F}{11},\frac{14F}{11}$
• B. $\frac{4F}{11},\frac{14F}{11}$
• C. $\frac{F}{11},\frac{14F}{11}$
• D. $\frac{3F}{11},\frac{14F}{11}$

Answer 1

The correct option is A $\frac{2F}{11},\frac{14F}{11}$

The forces, acceleration, angular acceleration of disc and rod are shown in the figure.

Equation for disc: $F^{\prime }\times r=\frac{3}{2}$ mr ${}^2\alpha$ - (i)

$F^{\prime }-f=m{a}_2-\left(ii\right)$

$a_2=r{\alpha }_2$ - (iii)

The equation of bar;

$P-P^{\prime }=\mathrm{ma}-\left(iv\right)$

$(F+F^{\prime })\frac{L}{2}=\frac{M{L}^2}{12}{\alpha }_1-\left(v\right)$

As point A on the box and disc are same,we have

$a_2=r{\alpha }_2=a-\frac{L}{2}{\alpha }_1-\left(vi\right)$

On solving (i) & (Vi), we get;

$a=\frac{8F}{11m},{\alpha }_1=\frac{24F}{22mL}$

$\left|a_A\right|=a_2=a-\frac{L}{2}{\alpha }_1=\frac{2F}{11m}$

$[a_B\mid =a+\frac{L}{2}{\alpha }_1=\frac{14F}{11m}$