A small object with a mass of m = 828 g is whirled at the end of a rope in a ver

Question

A small object with a mass of m = 828 g is whirled at the end of a rope in a vertical circle with a radius of r = 120 cm when it is at the location shown, (mid-height), its speed is v = 5.55 m/s. Determine the tension in the rope. (in N) 4.81 6.97 1.01x101 1.47x101 01 2.13x101 3.08x101 2.29 Submit Answer Tries 0/20 Calculate the magnitude of the total force acting on the mass at that location. 3.32 (in N) D: 7.46 |9.32 11.16x10| Submit Answer Tries 0/20 ||1.46×101 ||1.82x101 2.28 × 101 2.84 x 101 3.56x 101

Explanation / Answer

Solution:

Part A)Centripetal acceleration of mass m is given by

a = v^2 / r

Using Newton's 2nd law of motion, we have,
F = ma

substituting a
F = m v^2 / r ...........................(i)

Now,
m = 0.828 kg
r = 1.2 m

v = 5.55 m/s

using equation (i)
F = 0.828 * (5.55)^2 / 1.2

F = 21.3 N

Tension is equal to centripetal force here,
So option G is Right

Part B)

There are two forces acting on the mass , first is weight and 2nd is cetripetal force.

Weight (W) = mg = 0.828 x 9.8 = 8.1144 N

centripetal force = 21.3 N

Net fore = ( 21.3^2 + 8.1144^2)

Net Force = 22.7932 N

So, Option F is the Right one

Thanks. PLease comment for Queries.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.