A collar, constrained by a sliding pin to move on a circular track, has mass m -

Question

A collar, constrained by a sliding pin to move on a circular track, has mass m -5.8kg The circular track has a radius of R = 1.8m. The spring, which is connected to a pin that is fixed but can swivel, has a stiftness of k 100N/m. The spring's undeformed length is equal to the radius of the circular track (so the spring is in tension when the collar is at position 1) The colar is released from rest at position 1 and slides round to position 2 where = 430 Assume negligible friction, and g = 9.8m/s Recall that a circular track is described with respect to a port on its edge by 2 Recoas (o) Circular Track in Sliding pin and collar Fixed, swivel pin e,

Explanation / Answer

given, radius of track, R = 1.8 m

mass of collar, m = 5.8 kg

spring stiffness, k = 100 N/m

Now, the spring's natural length is R

so at position 1

gravitational PE of the system = mgR

Spring PE = 0.5kR^2

NEt energy = mgR + 0.5kR^2

at position 1

let the collar be moving with speed v

then KE = 0.5mv^2

PE = mg(R - Rsin(2*theta)) = mgR(1 - sin(2*theta)) [ by using the property that angle subtended by an arc of circle at the center is double the angle subtended at the edge]

and length of spring = 2R - R(1 - cos(2*theta))

so extension = 2R - R(1 - cos(2*theta)) - R = R(1 - 1 + cos(2*theta)) = Rcos(2*theta)

a. from conservation of eenrgy

0.5k*(Rcos(2*theta))^2 + mgR( 1 - sin(2(theta))) + 0.5mv^2 = mgR + 0.5kR^2

0.5k*(Rcos(2*theta))^2 - mgR*sin(2(theta))) + 0.5mv^2 = 0.5kR^2

now, theta = 43 deg

so putting in values

0.5*100*1.8^2*cos^2(43) - 5.8*9.81*1.8sin(2*43) + 0.5*5.8*v^2 = 0.5*100*1.8^2

v = 7.82 m/s

so the cvollar is moving at 7.82 m/s at point 2

b. r = Rcos(theta)

and v = d(2*theta)*R/dt

so, dr/dt = -Rsin(theta)*d(theta)/dt = -Rsin(theta)*v/2R

dr/dt = -sin(theta)*v/2 = -sin(43)*v/2

dr/dt at theta = 43 deg = -2.667 m/s

c. angular velocity d(theta)/dt = v/2R = 2.17222 rad/s

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