Question
A rock of mass $m=1.5 \mathrm{kg}$ is tied to a string of length $L=2.0 \mathrm{m}$ and is twirled in a vertical circle as shown in Eigure $5.10 .$ The speed $v$ of the rock is constant; that is, it is the same at the top and the bottom of the circle. If the tension in the string is zero when the rock is at its highest point (so that the string just barely goes slack), what is the tension when the rock is at the bottom?
Step 1
This means that the centripetal force, $mv^2/L$, is supplied by the weight of the rock, $mg$. So, we can write the equation of motion at the top as: \[T_A + mg = \frac{mv^2}{L}\] where $T_A$ is the tension at the top. Show more…
Show all steps
Your feedback will help us improve your experience
Saikat Chatterjee and 72 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
You are rotating a 2.0-kg rock in a vertical circle, using a very light string and such that the speed of the rock is constant. The radius of the circle is 0.50 m. When the rock is at the highest position, the tension in the string is 10 N. What is the tension in the string when the rock goes through the lowest position?
A rock of mass m= 1.50 kg, tied to the lower end of a string of length L from a ceiling, is set in uniform motion at speed v in a horizontal circle of radius r, What is the angle α between the string and the horizontal if r= 1.00 m and v= 3.00 m/s?( What is the tension on the string if r= 1.00 m and v= 3.00 m/s? What is the maximum value of v if L remains the same ( rand α can vary) and the maximum tension that the string can bear is 73.0 N?
A 5 kg object travels in a vertical circle of radius 10 m at constant speed of 15m/s. determine the tension in the string at the top of the circle.
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD