Question

example #12: angle
a 1.45-kg ball is suspended from a 0.8m string and swings in a horizontal circle at a constant speed such that the string makes an angle of 14 degrees with the vertical.
(a) draw an fbd of the mass.
(b) what is the tension in the string?
(c) what is the speed of the ball?
(d) if the ball had more mass, how would that affect the tension of the string?

Explanation:

Step1: Analyze vertical forces

In the vertical direction, the vertical component of tension \(F_{T}\) balances the weight of the ball. The weight \(W = mg\), where \(m = 1.45\space kg\) and \(g=9.8\space m/s^{2}\), so \(W=1.45\times9.8 = 14.21\space N\). The vertical component of tension is \(F_{T}\cos\theta\), where \(\theta = 14^{\circ}\). So \(F_{T}\cos\theta=mg\).

Step2: Solve for tension

We can solve for \(F_{T}\) from the equation \(F_{T}=\frac{mg}{\cos\theta}\). Substituting \(m = 1.45\space kg\), \(g = 9.8\space m/s^{2}\) and \(\theta=14^{\circ}\), \(\cos(14^{\circ})\approx0.97\). Then \(F_{T}=\frac{1.45\times9.8}{0.97}\approx14.94\space N\).

Step3: Analyze horizontal forces

In the horizontal direction, the horizontal component of tension \(F_{T}\sin\theta\) provides the centripetal force \(F_{c}=\frac{mv^{2}}{r}\). First, find the radius \(r = L\sin\theta\), where \(L = 0.8\space m\), so \(r=0.8\times\sin(14^{\circ})\approx0.19\space m\).

Step4: Solve for speed

We know \(F_{T}\sin\theta=\frac{mv^{2}}{r}\). Substitute \(F_{T}\approx14.94\space N\), \(\theta = 14^{\circ}\), \(m = 1.45\space kg\) and \(r\approx0.19\space m\). \(F_{T}\sin\theta=14.94\times\sin(14^{\circ})\approx3.62\space N\). Then from \(F_{c}=\frac{mv^{2}}{r}\), we have \(v=\sqrt{\frac{F_{c}r}{m}}=\sqrt{\frac{3.62\times0.19}{1.45}}\approx0.69\space m/s\).

Step5: Analyze effect of mass on tension

From \(F_{T}=\frac{mg}{\cos\theta}\), if \(m\) increases (while \(\theta\) remains the same as the speed - radius relationship would adjust accordingly in a physical sense for a conical pendulum, but in the formula), since \(F_{T}\) is directly proportional to \(m\) (with \(\cos\theta\) and \(g\) constant), the tension \(F_{T}\) would increase.

Answer:

(A) The free - body diagram (FBD) has two forces: the tension force \(F_{T}\) acting along the string and the weight force \(W = mg\) acting vertically downwards.
(B) \(F_{T}\approx14.94\space N\)
(C) \(v\approx0.69\space m/s\)
(D) The tension in the string would increase.