About Uniform circular motion,kindly solve this problem with solution.thanks!?

Question

An 8-lb ball attached to the end of a 3 ft long string is rotating as a conical pendulum. If it rotates at 60 rpm, What angle will the string make with the horizontal?

ans: 15.78 degree

Answer 1

First, figure out the ball's acceleration based on the description of its motion; next, apply "F=ma".

First, the "acceleration" part.  We know that ANYTHING traveling in a circle at constant speed, has this much acceleration:

a = ω²r

where:

ω = angular speed in radians/sec

r = radius of circle

Let's try to plug in some numbers for that.  To convert from "rpm" to "radians/sec", note that there are 2π radians in a revolution; and 60 seconds in a minute:

ω = 60 revs/minute × 2π rad/rev × (1 min/60 sec) = 2π rad/sec.

The radius of the circle can be calculated by knowing the length "L" of the rope (which we know); and its angle φ from the horizontal (which we don't (yet) know):

r = Lcosφ

(Notice that "φ" is the number we're trying to find as the answer).

So put these together:

a = ω²r = (2π rad/sec)²(Lcosφ)

Now hold that thought, and let's work on the "F=ma" part.

When you think about it, there are actually only two forces acting on the ball:

1. the tension in the rope (T), which pulls diagonally upward;

2. the force due to gravity (mg), which pulls straight down.

It helps to break the tension (T) into horizontal and vertical components:

Horizontal component: Tcosφ, pointing toward center of circle;

Vertical component: Tsinφ, pointing up.

Now we can break "F = ma" into two equations:

F_horizontal = (m)(a_horizontal)

F_vertical = (m)(a_vertical)

We know that F_horizontal = Tcosφ (all the horizontal force is provided by the rope); and we know a_horizontal = (2π rad/sec)²(Lcosφ) (from Part 1) so:

F_horizontal = (m)(a_horizontal)

Tcosφ = (m)(2π rad/sec)²(Lcosφ)

and conveniently, the "cosφ" cancels out of the above, so:

T = (m)(2π rad/sec)²(L) [Equation 1]

We know that F_vertical is the downward force (mg) minus the upward force (Tsinφ); and we know a_vertical = 0 (because the ball is not accelerating at all in the vertical direction); so:

F_vertical = (m)(a_vertical)

mg - Tsinφ = (m)(0) = 0

Now solve that for T:

T = mg/(sinφ) [Equation 2]

Now combine Equation 1 and Equation 2:

(m)(2π rad/sec)²(L) = mg/(sinφ)

And finally, solve that for "φ", which I'll leave up to you.  (Notice that the ball's mass "m" cancels out of the equation; so the fact that it is 8 lbs. is irrelevant to this problem.)

Answer 2

let me try to help you set this up...

draw a diagram of the weight on the end of the string...

there is tension T inthe string, and the vertical component of the tension will satisfy

the angle theta i am using is the angle between the string and the vertical, at the end we will take 90 minus this angle...

Tcos(theta)=mg (eq 1)

the horizontal component will be responsible for the centripetal acceleration, so we have:

Tsin(theta)=mv^2/r (eq 2)

we know m=1/4 slugs (in English units), and we need to find v and r

v=dist/time = 2 pi r/time

we know the pendulum makes 60 rpm, meaning it travels

60x2 pi r in 60s, or that v=2 pi r

but r is not 3 feet, rather r is 3 sin (theta)

so...to solve...

first divide eq(2) by eq (1) to eliminate T and m and get:

tan (theta) = v^2/rg

sub v=2 pi r into the eq above and get:

tan(theta)=4pi^2r/g

now sub r=3 sin(theta) to get:

tan(theta)=12 pi^2 sin(theta)/g

now remember tan=sin/cos, and solve to get:

1/cos(theta) = 12pi^2/g

using g=32.2 ft/s/s

and solving, you get theta=74.3 degrees from the vertical, or 15.3 from the horizontal