Question

A 220 g block on a 56.0 cm -long string swings in a circle on a horizontal, frictionless table at 50.0 rpm
a) find speed of the block
b) whats the tension in the string


thanks!

Answer 1

a) To find the speed of the block, we first have to convert the rpm to rad/s. We know that 1 revolution is equal to 2π radians, so the conversion factor is (2π rad/rev). We can then use the formula:

ω = 50.0 (rev/min) * (2π rad/rev) * (1 min/60 s)

Computing this as:

ω = 50.0 * (2 * 3.14159) * (1/60) = 5.236/s

Now that we have the angular velocity, we can find the linear velocity of the block by multiplying it by the radius of the circle:

v = r * ω

The radius is half of the string's length, which is 56.0 cm / 2 = 28.0 cm = 0.28 m. So we have:

v = 0.28 m * 5.236 s^-1 = 1.466 m/s

The speed of the block is 1.466 m/s.

b) To find the tension in the string, we will use the centripetal force equation:

Fc = m * v² / r

Where m is the mass of the block (220 g = 0.22 kg), v is the speed of the block (1.466 m/s), and r is the radius (0.28 m). Plugging these values into the equation, we get:

Fc = 0.22 kg * (1.466 m/s)² / 0.28 m = 1.497 N

The tension in the string is 1.497 N.