A small block of mass 200 grams moves with a uniform speed in a horizontal circulargroove. The radius of the groove’s vertical sidewalls is 50 cm. If the block takes 4.0 s tocomplete one round, find the normal force by the sidewall of the groove.
Question
A small block of mass 200 grams moves with a uniform speed in a horizontal circulargroove. The radius of the groove’s vertical sidewalls is 50 cm. If the block takes 4.0 s tocomplete one round, find the normal force by the sidewall of the groove.
Solution
To solve this problem, we need to first understand that the normal force exerted by the sidewall of the groove is the centripetal force that keeps the block moving in a circular path.
Step 1: Convert the given values to SI units. The mass of the block (m) = 200 grams = 0.2 kg The radius of the groove (r) = 50 cm = 0.5 m The time taken to complete one round (T) = 4.0 s
Step 2: Calculate the speed of the block. The speed (v) is given by the distance travelled divided by the time taken. In one round, the block travels a distance equal to the circumference of the circle, which is 2πr. So, v = 2πr/T = 2π*0.5/4.0 = 0.785 m/s
Step 3: Calculate the centripetal force. The centripetal force (F) is given by the formula F = mv²/r. Substituting the values we have, F = 0.2*(0.785)²/0.5 = 0.123 N
Step 4: Calculate the gravitational force. The gravitational force (Fg) is given by the formula Fg = mg, where g is the acceleration due to gravity. Taking g = 9.8 m/s², Fg = 0.2*9.8 = 1.96 N
Step 5: Calculate the normal force. The normal force (N) is the vector sum of the centripetal force and the gravitational force. Since these two forces are perpendicular to each other, we can use the Pythagorean theorem to find the resultant. So, N = sqrt(F² + Fg²) = sqrt((0.123)² + (1.96)²) = 1.97 N
So, the normal force exerted by the sidewall of the groove is approximately 1.97 N.
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