I rolled a 14 pound bowling ball at 23.36 MPH the other night and was wondering what the amount of force was that I used to roll it. (Considering it threw out my back and arm)
Is there an equation i can used to determine this?
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1 Answer
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The simplest way to do this is by invoking the concept of momentum $p=mv$, where $m$ is the ball's mass (in kilograms) and $v$ is its velocity (meters per second or miles per hour).
The ball rolls because it had an interaction with your arm. Newton's second law says that during the interaction, the increase (or change, denoted by $\Delta$) in the ball's momentum $p$ is equal to the average force $F$ applied by your arm, multiplied by the duration $t$ of the interaction, that is how long you spent rolling it. In equation form, $$ \Delta p = F t. $$ We can replace $\Delta p$ with $m \Delta v$, where $\Delta v$ is the increase in the ball's velocity during the interaction with your arm ($m$ doesn't change, but $v$ does). That gives $$ m \Delta v = F t.$$ Assuming the ball starts from rest (not moving), the increase in velocity will equal the final velocity, so we can replace $\Delta v$ with simply $v$, the ball's velocity after it has left your hand. That gives $$ mv = Ft. $$ Solving for $F$ gives $$ F=\frac{mv}{t}.$$ Now we just need to insert values for $m$, $v$, and $t$. The mass $m$ is easily converted from its weight: $$ m = 14\,\text{lbs} \times \frac{1\,\text{kg}}{2.2\,\text{lb}} = 6.4\,\text{kg}.$$ And we can convert the velocity to meters per second: $$ v = 23.36\,\text{mi/h} \times \frac{1600\,\text{m}}{1\,\text{mi}} \times \frac{1\,\text{h}}{3600\,\text{s}} = 10.4\,\text{m/s}.$$ We want to solve for the force $F$. We'll have to estimate $t$, the length of time that the ball was being pushed by your hand/arm. I'll just guess about half a second. You can apply your own estimate. Inserting my estimate: $$F = \frac{6.4\,\text{kg} \times 10.4\,\text{m/s}}{0.5\,\text{s}} = 133\,\text{kg m/s/s}.$$ The units of the answer are Newtons, which can be converted back to pounds as $$ 133\,\text{N} \times \frac{1\,\text{lb}}{4.45\,\text{N}} = 30\,\text{lb}.$$ So the average force applied to the ball during your roll was 30 pounds, if you believe my estimate of the duration $t$. Now there was a peak force that was possibly much higher than the average. And if you were applying that force with suboptimal technique, the local force on your muscles and joints could have been drastically higher in places.
