Question

A rectangular plate of glass, measured at rest, has sides \(30.0 \mathrm{cm}\) and \(60.0 \mathrm{cm} .\) (a) As measured in a reference frame moving parallel to the 60.0 -cm edge at speed \(0.25 c\) with respect to the glass, what are the lengths of the sides? (b) How fast would a reference frame have to move in the same direction so that the plate of glass viewed in that frame is square?

Short answer

$$L = (60.0 \mathrm{cm})\sqrt{1-\frac{(0.25)^2}{1}}$$ $$L = (60.0 \mathrm{cm})\sqrt{1-0.0625}$$ $$L = (60.0 \mathrm{cm})\sqrt{0.9375}$$ $$L \approx 58.08 \mathrm{cm}$$ So the length of the 60.0 cm side in the moving frame is approximately 58.08 cm. #tag_title#Determining the speed for the plate's sides to appear equal#tag_content#Now, we must find the speed for the plate's sides to appear equal in a moving frame. In this case, we want the contracted length of the 60.0 cm side to be equal to the 30.0 cm side. Using the length contraction formula: $$30.0 \mathrm{cm} = (60.0 \mathrm{cm})\sqrt{1-\frac{v^2}{c^2}}$$ Solve for \(v\): $$\frac{30.0 \mathrm{cm}}{60.0 \mathrm{cm}} = \sqrt{1-\frac{v^2}{c^2}}$$ $$\frac{1}{2} = \sqrt{1-\frac{v^2}{c^2}}$$ Square both sides: $$\frac{1}{4} = 1-\frac{v^2}{c^2}$$ Move the fractions: $$\frac{v^2}{c^2} = 1 - \frac{1}{4}$$ $$\frac{v^2}{c^2} = \frac{3}{4}$$ Multiply by \(c^2\) and take the square root: $$v = c\sqrt{\frac{3}{4}}$$ $$v = c\sqrt{0.75}$$ Therefore, the speed required for the plate's sides to appear equal in a moving frame is approximately 0.866 times the speed of light, or 0.866c.

Step-by-step solution

Calculating the length of the 60.0 cm side in the moving frame

In the moving frame, the length of the 60.0 cm side will be contracted. According to the length contraction formula: $$L = L_0\sqrt{1-\frac{v^2}{c^2}}$$ Where \(L\) is the contracted length, \(L_0\) is the proper length (60.0 cm), \(v\) is the relative speed, and \(c\) is the speed of light. Given the speed \(v = 0.25c\), we can substitute the values: $$L = (60.0 \mathrm{cm})\sqrt{1-\frac{(0.25c)^2}{c^2}}$$