A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction.

Question

A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction. (A) (3 R/2) (B) (5 R/2) (C) R (D) 2R

Tardigrade Admin · Accepted Answer

Key Point Apply conservation of energy at two different position of the object.

According to energy conservation between points

A

and

B

m g (2 R) + \frac{1}{2} (0)^{2} = m g H + \frac{1}{2} m (0)^{2}

\Rightarrow m g (2 R) = m g H

\Rightarrow 2 R = H

\Rightarrow H = 2 R

Q. A small bead of mass $M$ slides on a smooth wire that is bent in a circle of radius $R$ . It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction.

$\frac{3 R}{2}$

$\frac{5 R}{2}$

$R$

$2 R$

Key Point Apply conservation of energy at two different position of the object.

According to energy conservation between points $A$ and $B$
$m g (2 R) + \frac{1}{2} (0)^{2} = m g H + \frac{1}{2} m (0)^{2}$
$\Rightarrow m g (2 R) = m g H$
$\Rightarrow 2 R = H$
$\Rightarrow H = 2 R$

Q. A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction.

23R​

25R​

R

2R

Key Point Apply conservation of energy at two different position of the object. According to energy conservation between points A and B mg(2R)+21​(0)2=mgH+21​m(0)2 ⇒mg(2R)=mgH ⇒2R=H ⇒H=2R

Q. A small bead of mass $M$ slides on a smooth wire that is bent in a circle of radius $R$ . It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction.

$\frac{3 R}{2}$

$\frac{5 R}{2}$

$R$

$2 R$

Key Point Apply conservation of energy at two different position of the object.

According to energy conservation between points $A$ and $B$
$m g (2 R) + \frac{1}{2} (0)^{2} = m g H + \frac{1}{2} m (0)^{2}$
$\Rightarrow m g (2 R) = m g H$
$\Rightarrow 2 R = H$
$\Rightarrow H = 2 R$