Here's another one that can't seem to figure out.?

Question

A 2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is theta =30 degrees?What is the ball's speed when theta =30 degrees? What is the ball's magnitude of acceleration at theta =30 degrees?

Answer 1

Ft = mgsin(30) = ma
a = gsin(30)
a = 9.8*0.5 = 4.9ms^-2 at the angle of 30 deg
Now T + mgcos(30) = mv^2/R
where mv^2/R is the centrifugal force
solving the above equation by plugging in the values for T, m, and R, we get
20N + 2.0*9.8*0.87 = 2.0*v^2/0.8
37.05 =2.5v^2
Hence v^2 = 37.05/2.5 = 14.82
and v = sqrt(14.82) =3.85m/s

Answer 2

If the highest point is at 30 degrees, that is the point at which the ball stops moving up and begins to move down. At that instant the speed must = zero. The only force is the force of gravity which is 9.8m/s^2 straight down. I'm not sure how to convert for the 30 degree angle, but it shouldn't be difficult.

Answer 3

V = a million/3 Bh So, all of us understand that V = 6, and h = 2 as a result, 6=a million/3 * B * 2 3= a million/3 * B B = 9 as a result, the portion of the backside is 9 sq. centimeters. we are in a position to bypass further, when you consider that, the backside is a circle, B = pi *r^2 9 = pi *r^2 r^2 = 9/pi r = root 9/pi = 3/root pi cm