Question

Grace, playing with her dolls, pretends the turntable of an old phonograph is a merry-go-round. The dolls are \(12.7 \mathrm{cm}\) from the central axis. She changes the setting from 33.3 rpm to 45.0 rpm. (a) For this new setting, what is the linear speed of a point on the turntable at the location of the dolls? (b) If the coefficient of static friction between the dolls and the turntable is \(0.13,\) do the dolls stay on the turntable?

Short answer

(b) Will the dolls stay on the turntable given the coefficient of static friction as 0.13?

Step-by-step solution

(a) Convert rotational speed to rad/s

First, we need to convert the rotational speed in rpm to radians per second (rad/s). The conversion factor is 2π radians per rotation and 1 minute per 60 seconds: \(ω = 45.0 \, \frac{\text{rpm}}{\text{min}}\times \frac{2π\, \text{rad}}{\text{rotation}} \times \frac{1\, \text{min}}{60\, \text{s}}\)

Calculate the linear speed

Now, we can calculate the linear speed at the location of the dolls using the formula \(v = ωr\), where \(v\) is the linear speed, \(ω\) is the rotational speed (in rad/s), and \(r\) is the distance from the central axis. \(v = ωr = (45.0 \, \frac{\text{rpm}}{\text{min}} \times \frac{2π\, \text{rad}}{\text{rotation}} \times \frac{1\, \text{min}}{60\, \text{s}}) \times (12.7\, \mathrm{cm} \times \frac{1\, \text{m}}{100\, \text{cm}})\)

(b) Calculate the centripetal acceleration

Next, we'll calculate the centripetal acceleration using the formula \(a_c = ω^2r\), where \(a_c\) is the centripetal acceleration, \(ω\) is the rotational speed (in rad/s), and \(r\) is the distance from the central axis. \(a_c = ω^2r = (45.0 \, \frac{\text{rpm}}{\text{min}} \times \frac{2π\, \text{rad}}{\text{rotation}} \times \frac{1\, \text{min}}{60\, \text{s}})^2 \times (12.7\, \mathrm{cm} \times \frac{1\, \text{m}}{100\, \text{cm}})\)

Calculate the maximum static friction

We are given the coefficient of static friction as \(μ_s = 0.13\). We can calculate the maximum static friction force using \(F_{max} = μ_s m g\), where \(F_{max}\) is the maximum static friction force, \(m\) is the mass of the dolls, and \(g\) is the gravitational acceleration (9.81 m/s²). We don't know the mass of the dolls, so let's express the maximum static friction force in terms of the mass: \(F_{max} = 0.13 m g\)

Compare the centripetal force to the maximum static friction

Finally, we need to compare the centripetal force \(F_c\) to the maximum static friction force to determine if the dolls stay on the turntable or not. The centripetal force is given by \(F_c = ma_c = mω^2r\). If the centripetal force is less than or equal to the maximum static friction force, the dolls will not slip: \(mω^2r \leq 0.13 mg\) Dividing both sides by \(m\), we get: \(ω^2r \leq 0.13 g\) Now, we can plug in the values for \(ω\), \(r\), and \(g\) to check if this inequality holds true. If it does, the dolls will stay on the turntable; if not, they will slip.