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

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