A turntable with a moment of inertia I = 0.25 kg m² is initially spinning at an angular velocity of ω₀ = 2.3 rad/s. If friction in the bearings exerts a constant torque of τ = 0.28 Nm, how long does it take for the turntable to stop spinning?
Added by Emily F.
Step 1
First, we need to understand that the torque exerted by friction is acting to slow down the turntable. This means it is acting in the opposite direction to the initial spin, so we can consider it as a negative torque. Show more…
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A turntable with a moment of inertia I = 0.25 kg m^2 is initially spinning at an angular velocity of ω0 = 6.5 rad/s. If friction in the bearings exerts a constant torque of τ = 0.15 Nm, how long does it take for the turntable to stop spinning? Enter your answer in seconds, without the unit. Hint: This problem is conceptually similar to that of a car moving at an initial speed of v0 and is slowed down by a constant force F. For full credit, your answer must be within ±2% of the correct value.
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The moment of inertia of the empty turntable is $1.5 \mathrm{~kg} \mathrm{~m}^{2}$. With a constant torque of $2.5 \mathrm{~N} \cdot \mathrm{m},$ the turntable-person system takes $3.0 \mathrm{~s}$ to spin from rest to an angular speed of $1.0 \mathrm{rad} / \mathrm{s}$. What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) $2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) $6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (c) $7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (d) $9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}$.
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