A stone of mass 50 grams is attached to one end of a string of length 1 meter The maximum tension in the string can withstand is 200N . The stone is rotated uniformly in a horizontal circle. Read the above passage and answer the following questions.

Given that a stone with a mass of 50 grams is attached to a 1-meter string and spun in a horizontal circle, if the maximum tension the string can withstand is 200N, what is the maximum speed at which the stone can be rotated without breaking the string?

Solution:

To find the maximum speed at which the stone can be rotated without breaking the string, we can use the formula for the centripetal force, which must be equal to or less than the maximum tension the string can withstand. The formula for centripetal force

F_c

F_{c}

is:

F_c = \frac{m v^2}{r}

F_{c} = r m v ^{2}

Where:

$m$ $m$ is the mass of the stone,
$v$ $v$ is the velocity (speed) of the stone,
$r$ $r$ is the radius of the circle (length of the string).

Given:

$m = 50$ $m = 50$ grams = 0.05 kg (since $1 \text{ gram} = 0.001 \text{ kg}$ $1 gram = 0.001 kg$ ),
$r = 1$ $r = 1$ meter,
The maximum tension (maximum force $F_c$ $F_{c}$ ) the string can withstand = 200 N.

We rearrange the formula to solve for

v

v

v = \sqrt{\frac{F_c \cdot r}{m}}

v = m F ^{c} \cdot r

Substituting the given values:

v = \sqrt{\frac{200 \cdot 1}{0.05}}

v = 0.05200 \cdot 1

Let's calculate that.

The maximum speed at which the stone can be rotated without breaking the string is approximately

63.25 \, \text{m/s}

63.25 m/s

A stone of mass 50 grams is attached to one end of a string of length 1 meter The maximum tension in the string can withstand is 200N . The stone is rotated uniformly in a horizontal circle. Read the above passage and answer the following questions.

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us