Question

An object of mass 500g is attached to a string of length 50cm which will break if the tension is 20N . The object is whirled in a vertical circle , the axis of rotation being at a height of 100cm above the ground. The angular speed is very slowly increased until the string breaks.In what position is this breakage likely to occur, and of what angular speed? Where will the object hit the ground?

Answer 1

Answer on Question #64840 – Physics – Mechanics/Relativity

Question: An object of mass 500g is attached to a string of length 50cm which will break if the tension is 20 N. The object is whirled in a vertical circle, the axis of rotation being at a height of 100cm above the ground. The angular speed is very slowly increased until the string breaks. In what position is this breakage likely to occur, and of what angular speed? Where will the object hit the ground?

Solution: There are three forces acting on an object: gravity ($F_g$, directed vertically), centrifugal force ($F_c$, directed away from the axis of rotation) and tangential force ($F_t$, directed tangent to the circle). The string will break when the resultant force will exceed 20 N.

In the first approximation we can consider that tangential force tends to zero because $F_t = m\beta R$, where $m$ is the body's mass, $\beta$ is the angular acceleration and $R$ is the circle's radius. But the angular acceleration tends to zero because in the system "the angular speed is very slowly increased". However, the strings will break when the gravity's vector and centrifugal forces' vector will collinear and the resultant force will exceed 20 N. Accord to the fact that gravity's vector always directed vertically, the only point where two vectors are collinear is the lowest point of circle. At this point the total force ($F$) is equal to: $F = F_g + F_c = mg + m\omega^2 R$, where $m$ is the body's mass (in kg), $g$ is the gravitational acceleration (about 9.8 m/s²), $\omega$ is the angular speed (in rad/s) and $R$ is the circle's radius (in m). So, the angular speed is equal to: $\omega = \sqrt{\frac{F}{mR} - \frac{g}{R}}$. The body's mass is 0.5 kg, the circle's radius is 0.5 m. Calculate: $\omega = \sqrt{\frac{20N}{0.5kg*0.5m} - \frac{9.8m}{0.5ms^2}} = \sqrt{80 - 19.6} = 7.77s^{-1}$.

After the string's breaking an object continues to move in two independent directions. The first direction is collinearly to gravity's vector. The vertical way that an object moves after a string is break is 50 cm (a height of rotational axis (100 cm) minus a circle's radius (50 cm). In this case the object's movement described by this equation: $h = \frac{gt^2}{2}$, where $t$ is the movement's time (in s). Therefore, the object movement's time is $t = \sqrt{\frac{2h}{g}} = \sqrt{0.1} = 0.316s$.

The second direction is horizontally because at the moment of string's breaking the object moves in this direction with some velocity ($v$) and continues to move by inertia. The velocity depends on the angular speed: $v = \omega R = 7.77 * 0.5 = 3.885 \, m/s$. At this speed horizontal way is: $s = v * t = 3.885 * 0.316 = 1.228 \, m$. The object hits the ground at a distance of 1.288 m from the perpendicular from circle's center point to the ground in the plane that circle forms.

Answer: The string will break when the object will be at the lowest point of circle with the angular speed of $7.77 \, \text{s}^{-1}$. After string's breaking the object hits the ground at a distance of $1.288 \, \text{m}$ from the perpendicular from circle's center point to the ground in the plane that circle forms.

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