Question

The figure shows two wires that are tied to a 710 g mass that revolves in a horizontal circle at a constant speed of 7.5 m/s. (a) What is the tension in the upper wire? (b) What is the tension in the lower wire?(c) If the max tension in a wire can only be 30N. What would happen in this situation? What would be the new tension in order to maintain a speed of 7.5 m/s?

Answer 1

Answer:

(a). Upper wire = 33.59 N

(b). downward wire = 19.67 N

(c). The upper wire would break.

Explanation:

The centripetal force  that keeps the mass moving in a circular path is directed towards the center and has the magnitude

$F =\dfrac{M*v^2}{R}$

Since

$R = \sqrt{1.0m^2-0.5m^2} = \dfrac{\sqrt{3} }{2}$

$m= 0.710kg$,

and $v= 7.5m/s$

the centripetal force is

$F =\dfrac{0.710(7.5)^2}{\dfrac{\sqrt{3} }{2} }$

$F= 46.12N$

The two wires are supporting this force; therefore,

$T_1 cos(\theta)+T_2 cos(\theta)=46.12N$.

$(1).\:\: (T_u+T_d)cos(\theta)=46.12N$

Also, gravity is pulling down on the mass, and the wires must also support that; therefore,

$T_usin(\theta)= mg+T_dsin(\theta)$

$(2).\: \: (T_1-T_2)sin(\theta)= mg$

Now,

$\theta = sin^{-1}(\dfrac{0.5m}{1.0m} )= 30^o$;

therefore, equations (1) and (2) give

$(T_1+T_2)=\dfrac{46.12N }{cos(30^o)}$

$\boxed{(3). \:(T_1+T_2)=53.25N}$

and

$(T_1-T_2)= \dfrac{mg}{sin(\theta)}$

$\boxed{(4). \: T_1-T_2 = 13.916}$

Solving equations (3) and (4) gives

$\boxed{T_1= 33.59N}\\\\\boxed{T_2 = 19.67N}}$

Thus, the tension in the upper wire is 33.59 N and in the bottom wire 19.67 N.

(c).

If the maximum tension in a wire can only be 30 N, then the upper wire would break because we see that its tension of 33.59 N exceeds 30 N.