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Inertia problem

In Fig. 11-42, one block has a mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 81.5 cm in 2.90 s (without the cord slipping on the pulley).

(a) What is the magnitude of the block's acceleration? _______m/s2
(b) What is the tension in the part of the cord that supports the heavier block? _______N
(c) What is the tension in the part of the cord that supports the lighter block? _______N
(d) What is the magnitude of the pulley's angular acceleration?
_______rad/s2
(e) What is its rotational inertia?
_______kg m2

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Solution Preview

Step 1.
The general equation relating constant acceleration a, to position y, time t, and initial velocity vo, is:
(1) y= vo t + .5 a t^2 in which vo=0.

Step 2.
In (1), knowing y and t and vo, you should substitute knowns to find:
(a) ANSWER acceleration (2) a= .194 nt/kg or .194 m/sec^2

Step 3.
On the ...

Solution Summary

This solution is provided in 300 words and gives step-by-step equations to find each of the answers to the missing information from the question.

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