We are constructing a last chance fire exit involving jumping off of a building tied to a cable that is wrapped around a barrel.

In this design the person's mass (m) is 85 kg. The barrel mass (M) is 985 kg and has a radius (R) of 1.5 meters, and the building height (h) is 12 meters.

Potential energy before the jump equals kinetic energy just before touchdown, so...

mgh = 0.5mv2 +0.5(0.5MR2((v/R)2

What is the landing velocity?

Question

We are constructing a last chance fire exit involving jumping off of a building tied to a cable that is wrapped around a barrel.

In this design the person's mass (m) is 85 kg. The barrel mass (M) is 985 kg and has a radius (R) of 1.5 meters, and the building height (h) is 12 meters.

Potential energy before the jump equals kinetic energy just before touchdown, so...

mgh = 0.5mv2 +0.5(0.5MR2((v/R)2

What is the landing velocity?

BlackTom AI · Accepted Answer

To begin, let's identify the energy balance given in the problem: the potential energy before the jump equals the kinetic energy just before touchdown. This gives mgh = K, where K is the total kinetic energy of the system just before landing.

Next, compute the total kinetic energy. The velocity of the person is v, and the barrel (with mass M and radius R) contributes rotational kinetic energy about its axis. The translational kinetic energy of the person is 0.5 m v^2. The barrel, assuming it rolls without slipping around the same axis, has rotational kinetic energy 0.5 I (v/R)^2, where I is the moment of inertia of the barrel.

For a solid cylinder (barrel) of mass M and radius R, the moment of inertia about its axis is I = 0.5 M R^2. Substituting into the rotational term yields 0.5 I (v/R)^2 = 0.5 (0.5 M R^2) (v^2 / R^2).

Now simplify the rotational contribution: 0.5 × 0.5 M R^2 × (v^2 / R^2) = 0.25 M v^2. Therefore, the total kinetic energy is K = 0.5 m v^2 + 0.25 M v^2 = v^2 (0.5 m + 0.25 M).

With these expressions in place, substitute m = 85 kg, M = 985 kg, g ≈ 9.8 m/s^2, and h = 12 m into the energy balance: mgh = v^2 (0.5 m + 0.25 M).

Compute the left side: mgh = 85 × 9.8 × 12 = 9996 (units: J).
Compute the factor in parentheses on the right: 0.5 m + 0.25 M = 0.5 × 85 + 0.25 × 985 = 42.5 + 246.25 = 288.75.

Thus v^2 = (mgh) / (0.5 m + 0.25 M) = 9996 / 288.75 ≈ 34.6 (m^2/s^2).

Finally, take the square root to obtain the landing velocity: v ≈ sqrt(34.6) ≈ 5.88 m/s (approximately 5.89 m/s when rounded to two decimal places).

Key checks: ensure the rotational term was correctly derived from I = 0.5 M R^2 and the (v/R)^2 factor, and verify the arithmetic for the sums in the denominator and the energy values to avoid sign or unit errors.

ENGR-216:501,502,503,504,505,506,507,508,509,510,511,512,513,514,515,PHYS-216:501,502,503,504,505,506,507,508,509,510,511,512,513,514,515,516,517,518,519,520,521,522,523,524 Final Exam Prep Quiz

View Explanation

Log in for full answers

Similar Questions

What is a rotational kinetic energy and what is its SI unit?

An ice skater performs a fast spin by pulling outstretched arms down and close to her body. What happens to her kinetic energy with respect to the axis of rotation?

Suppose a solid uniform cylinder rolls without slipping down an inclined plane (like in Figure 12.42). Once the object starts rolling then the rotational kinetic energy of the cylinder

A system consists of a 2.0 kg uniform disk with radius 0.40 m, and ring of negligible thickness and mass 1.0 kg with a radius of 0.20 m, as shown. If the system has a rotational speed of 50.0 rad/s, what is its kinetic energy in Joules?

In a consumer society, many adults channel creativity into buying things

Economic stress and unpredictable times have resulted in a booming industry for self-help products

People born without creativity never can develop it

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler