Question

6.) A meter stick of uniform density is balanced horizontally on a fulcrum with support at its center of mass (i.e. the 50 cm mark). A weight is hung at the 35 cm mark and the meterstick begins to rotate with an angular acceleration of 0.20 rad/s^2. Given that the moment of inertia of the meterstick is (1/12)*M*L^2 and the mass M of the meter stick is 0.12 kg A) What is the torque on the meter stick?
B.) What is the amount of weight hung at the 35 cm mark?
C.) Does the torque increase, decrease or remain constant as the meter stick begins to rotate?

Answer 1

Using law of rotate

$$T=J\epsilon$$

Moment of inertia with weight

$$J=\frac{1}{12}ML^2+mr^2$$ $$r=0.15 (m)$$

Weight is

$$P=mg$$

and torque

$$T=Pr$$

Find the mass

$$m=\frac{1}{12}\frac{ML^2\epsilon}{(g-\epsilon r)r}$$

$$m=\frac{1}{12}\frac{0.12\cdot 1^2\cdot 0.2}{(9.8-0.2\cdot 0.15)\cdot 0.15}=1.36 (g)$$

1)Torque

$$T=0,00136\cdot 9.8\cdot 0.15 = 2 (mN\cdot m)$$

2)Weigth

$$P=0.00136\cdot 9.8=13.37 (mN)$$

3)The torque decrease