Example 1.3 - Chapter 1 - Units and Measurement

Transcript

Example 1.3 Let us consider an equation (1)/(2)mv² = mgh, where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct. We need to compare dimensions of left side and right side. Let’s first write dimensions of each term individually Now, Dimension of mass m = [𝑀] Dimension of velocity v = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒)/(𝑇𝑖𝑚𝑒) = [(𝐿)/(𝑇)]= [𝐿𝑇^(−1)] Dimension of acceleration due to gravity g = (𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦)/(𝑇𝑖𝑚𝑒) = [(𝐿𝑇^(−1))/(𝑇)] = [𝐿𝑇^(−2)] Dimension of height h = [𝐿] Our equation is (𝟏)/(𝟐)mv² = mgh Since Dimensions of LHS = Dimensions of RHS Thus, our equation is dimensionally correct Dimensions of LHS Dimension of (1)/(2)mv² = [𝑀] × [𝐿𝑇^(−1)]^(2) = [𝑀] ×[𝐿^(2)𝑇^(−2)] = [𝑴𝑳^(𝟐)𝑻^(−𝟐)] Dimensions of RHS Dimension of mgh = [𝑀] × [𝐿𝑇^(−2)] × [𝐿] = [𝑴𝑳^(𝟐)𝑻^(−𝟐)]