Chapter 8.1, Problem 4PP

a

To determine

To Compare: The linear acceleration of trailer and truck.

Explanation of Solution

Given:

Diameter of trailer wheels, $d=48\text{cm}=0.48\text{m}$

Linear acceleration of truck, $a=1.85{\text{m/s}}^{\text{2}}$

Angular acceleration of truck, $\alpha =5.3{\text{rad/s}}^{\text{2}}$

Diameter of truck’s wheels, $d_t=0.7\text{m}$

Calculations:

As the truck is towing the trailer, they will have same in velocity per unit time i.e. they will have same linear acceleration.

Conclusion:

Thus, the linear acceleration of trailer and truck would be equal.

b

To determine

To Compare: The angular acceleration of trailer and truck.

Explanation of Solution

Given:

Diameter of trailer wheels, $d=48\text{cm}=0.48\text{m}$

Linear acceleration of truck, $a=1.85{\text{m/s}}^{\text{2}}$

Angular acceleration of truck, $\alpha =5.3{\text{rad/s}}^{\text{2}}$

Diameter of truck’s wheels, $d_t=0.7\text{m}$

Formula Used:

Angular acceleration can be obtained by:

  $\alpha =\frac{a}{r}$

Where, a is the linear acceleration and r is the radius.

Calculations:

The linear acceleration of truck and trailer are equal. Therefore, the ratio of angular acceleration of trailer can be calculated as follows:

  $\frac{{\alpha }_{trailer}}{{\alpha }_{truck}}=\frac{r_{truck}}{r_{trailer}}$

Multiply numerator and denominator by 2.

  $\frac{{\alpha }_{trailer}}{{\alpha }_{truck}}=\frac{2r_{truck}}{2r_{trailer}}=\frac{d_t}{d}$

Substitute the known values and solve:

  $\begin{array}{c}\frac{{\alpha }_{trailer}}{{\alpha }_{truck}}=\frac{d_t}{d}\\ =\frac{0.7}{0.48}\\ {\alpha }_{trailer}=1.46{\alpha }_{truck}\end{array}$

Conclusion:

Thus, the angular acceleration of trailer is 1.46 times the angular acceleration of the truck.