A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Asked by Siva 15/12/2021 Last Modified 13 Apr

1 Answer

Learn Unit 3-Laws of Motion

Follow 1

Answer

Please enter your answer

Question

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Asked by Siva 15/12/2021 Last Modified 13 Apr

1 Answer

Learn Unit 3-Laws of Motion

Follow 1

Answer

Nazia Khanum · Accepted Answer

As an experienced tutor registered on UrbanPro, I can confidently guide you through this physics problem. When dealing with circular motion, such as in this scenario, it's crucial to understand the forces acting on the object.

First, let's find the tension in the string when the stone is whirled at a speed of 40 revolutions per minute (rev/min) in a horizontal plane with a radius of 1.5 meters.

We'll start by converting the speed from rev/min to radians per second, as it's a more suitable unit for our calculations.

Given:

Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Speed, v = 40 rev/min

To convert rev/min to radians per second, we use the formula:

angular velocity(ω)=2π×revolutions60×secondsangular velocity(ω)=60×seconds2π×revolutions

Plugging in the values:

ω=2π×4060≈4.1888 rad/sω=602π×40≈4.1888rad/s

Now, we can find the tension in the string using the centripetal force formula:

Tension=mass×centripetal accelerationTension=mass×centripetal acceleration

Centripetal acceleration=v2rCentripetal acceleration=rv2

Given that v=ω×rv=ω×r, we substitute this into the equation:

Centripetal acceleration=(ω×r)2r=ω2×rCentripetal acceleration=r(ω×r)2=ω2×r

Substituting the values we found earlier:

Centripetal acceleration=(4.1888)2×1.5≈27.7734 m/s2Centripetal acceleration=(4.1888)2×1.5≈27.7734m/s2

Now, we can find the tension:

Tension=0.25×27.7734≈6.94335 NTension=0.25×27.7734≈6.94335N

So, the tension in the string when the stone is whirled at a speed of 40 rev/min is approximately 6.94 N.

Next, let's determine the maximum speed at which the stone can be whirled around if the string can withstand a maximum tension of 200 N.

Given:

Maximum tension, Tmax=200 NTmax=200N

We'll use the same centripetal force formula:

Tension=mass×centripetal accelerationTension=mass×centripetal acceleration

And solving for the maximum speed (vmaxvmax):

vmax=Tmaxm×rvmax=mTmax×r

Plugging in the values:

vmax=2000.25×1.5=200×1.5≈300≈17.32 m/svmax=0.25200×1.5

=200×1.5≈300

≈17.32m/s

So, the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N is approximately 17.32 m/s.

Feel free to ask if you have any questions or need further clarification!

I am a Student I am a Tutor
Name*	Please enter your full name. Please enter institute name.
Email*	Please enter your email address.
Phone*	Please enter a valid phone number.
Location*	Please enter a pincode or area name.
City*	Please enter city name.
Category*	Please enter category.
Gender*	Male Female Please select your gender.
Email ID/ Mobile No.*	Please enter either mobile no. or email.
Enter Password*	Please enter OTP Please enter Password Sorry, this phone number is not verified, Please login with your email Id.

Looking for Class 11 Tuition Classes?

Take Class 11 Tuition with the Best Tutors