Question

2. A 100 cm rod is free to pivot about one end, and a 50 N force is applied at the other end at an angle 20 degree to the horizontal.

(i) Calculate the lever arm of the rod.

(ii) Calculate the torque about the pivot point.

3. Joe weighing 500 N sits 0.5 m to the left of center of the seesaw 2 m long. Liz sits at the end on the opposite side of a seesaw of mass 50 kg such that the system is in equilibrium.

(i) Find the weight of Liz

(ii) Find the normal force acting at the pivot point.

4. A car is driving at constant angular speed along a 10 m radius circular track. The car completes one revolution of the track in 2 seconds. Calculate,

(i) The tangential speed of the car

(ii) The angular speed of the car.

(iii) The angle spanned by the car in 2 seconds.

(iv) The centripetal acceleration of the car.

5. A wheel rotates with a constant angular velocity of 3.5 m/s^2. If the angular speed of the wheel is 2 rad/s at t = 0 s,

(i) Find the angle through which the wheel rotates between t = 2.0 s to t = 3.0 s?

(ii) Find the angular speed when t = 3.0 s .

6. A car slows down uniformly from 50 m/s to 20 m/s in 2 s while moving in a circular path of radius of 100 m. When the car approaches a speed of 40 m/s, Find the car’s,

(i) centripetal acceleration

(ii) angular speed

(iii) tangential acceleration

(iv) total acceleration

Answer 1

2) L = 100 cm = 1 m ; F = 50 N ; theta = 20 deg

a)Using trigonometry, the lever arm will be:

d = 1 x sin20 = 0.342 m

Hence, d = 0.342 m

b)torque is:

tau = F x d

tau = 50 x 0.342 = 17.1 Nm

Hence, torque = 17.1 N-m

4) R = 10 m ;

(i)v = d/t

d = 2 pi r = 2 x 3.14 x 10 = 62.8 m

v = 62.8/2 = 31.4 m/s

v = 31.4 m/s

(ii)v = r w => w = v/r

w = 31.4/10 = 3.14 rad/s

w = 3.14 rad/s

(iii)one revolution = 360 deg

angle = 360 deg

(iv)ac = v^2/R = 31.4^2/10 = 98.6 m/s^2

Hence, ac = 98.6 m/s^2