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more qus

1. A spin drier tub operates at 800rpm. The tub has a diamter of 0.80m. calculate (a) The speed of a point on the tub wall
(b) The centripetal acceleration of the tub wall

2. An object of mass 0.30kg is attached to the end of a string and is supported on a smooth horziontal surface. The object moves in a horizontal circle of radius 0.50m with a constant speed of 2ms-1. Cacluate (a) the centripetal acceeleration and the tension in the string.

i wud appreciate it if some1 helped. thanks.

Reply 1

JayEm

a)1 rpm = pi/30 rads^-1
So 800 rpm = 80pi/3 rads^-1

v=wr
Therefore v=80pi/3 x 0.4

b) a=v²/r, just plug the numbers in, v which you calculated in part a) and the radius r.

what about qu 2 m8?

potter

2. An object of mass 0.30kg is attached to the end of a string and is supported on a smooth horziontal surface. The object moves in a horizontal circle of radius 0.50m with a constant speed of 2ms-1. Cacluate (a) the centripetal acceeleration and the tension in the string.

i wud appreciate it if some1 helped. thanks.

a) a = v^2/r
b) T = ma = mv^2/r

This is really tedious. Why do you need help? No offence.

Reply 4

JayEm

Yeah for question 2, the acceleration is just using v²/r. For the tension use T=mv²/r so basically you're multiplying the acc. by the mass.

Reply 5

JayEm

Just plug the values in for each term in the equations.

Reply 6

Mr power

i nee dhelp becos i find it ard. thanks guys.

Reply 7

JayEm

Get yourself a text book with a load of examples in and answers in the back. When I did M1, I didn't understand a thing in class, I taught myself all of it. If you need anymore help just post here. :smile: