Question

A point object of mass m is slipping down on a smooth hemispherical body of mass M and radius R. The point object is tied to a wall by an ideal string as shown. At a certain instant shown in figure, speed of the hemisphere is $v$ and its acceleration is a. Then speed $v_p$ and acceleration $a_p$ of the point object is (Assume all the surfaces in contact are frictionless).

• A. $v_p=\frac{v\sqrt{3}}{2}$
• B. $v_p=v$
• C. $a_p=a$
• D. $a_p=\sqrt{{(\frac{a\sqrt{3}}{2}+\frac{v^2}{R})}^2+{\left(\frac{a}{2}\right)}^2}$

Answer 1

Answer: (B), (D)

The correct options are
B $v_p=v$
D $a_p=\sqrt{{(\frac{a\sqrt{3}}{2}+\frac{v^2}{R})}^2+{\left(\frac{a}{2}\right)}^2}$


Length of rope is constant.
Therefore, $x+R\theta =constant$
$\frac{dx}{dt}+\frac{Rd\theta }{dt}=0$
as $\frac{dx}{dt}=v$
$\therefore \omega =\frac{-d\theta }{dt}=\frac{v}{R}$
Now, $\alpha =\frac{d\omega }{dt}=\frac{a}{R}$

Now, for resultant velocity, we will find relative velocity of mass m in frame of hemisphere and then superimpose hemisphere velocity to find resultant velocity of mass m.
At $\theta ={60}^{\circ }$, $v_p$ will be resultant of $v$ because of linear velocity of hemisphere and $r\omega =v$ because of slipping tangential to hemisphere.
Since, magnitude of both the vectors are same and angle between them is ${120}^{\circ }$ therefore, $v_p=v$.

In frame of hemisphere, mass m is slipping and moving along the circle. Therefore, net acceleration of mass m $a_p$ will be resultant of $a$ because of linear acceleration of hemisphere and $r\alpha =a$ because of slipping tangential to hemisphere and a centripetal acceleration $(\frac{v^2}{R}$) radial to hemisphere.
Resolving these acceleration in radial and tangential direction, we get $a_r=\frac{a\sqrt{3}}{2}+\frac{v^2}{R}$ and $a_t=a-\frac{a}{2}=\frac{a}{2}$
$a_p=\sqrt{{\left(a_r\right)}^2+{\left(a_t\right)}^2}$
$\Rightarrow a_p=\sqrt{{(\frac{a\sqrt{3}}{2}+\frac{v^2}{R})}^2+{\left(\frac{a}{2}\right)}^2}$